Solve for a a^2+b^2+2a-6b+10=0

$a^{2} + b^{2} + 2 a - 6 b + 10 = 0$

Step 1

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 2

Substitute the values $a = 1$ , $b = 2$ , and $c = b^{2} - 6 b + 10$ into the quadratic formula and solve for $a$ .

$\frac{- 2 \pm \sqrt{2^{2} - 4 \cdot (1 \cdot (b^{2} - 6 b + 10))}}{2 \cdot 1}$

Step 3

Simplify.

Simplify the numerator.

Raise $2$ to the power of $2$ .

$a = \frac{- 2 \pm \sqrt{4 - 4 \cdot 1 \cdot (b^{2} - 6 b + 10)}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$a = \frac{- 2 \pm \sqrt{4 - 4 \cdot (b^{2} - 6 b + 10)}}{2 \cdot 1}$

Apply the distributive property.

$a = \frac{- 2 \pm \sqrt{4 - 4 b^{2} - 4 (- 6 b) - 4 \cdot 10}}{2 \cdot 1}$

Simplify.

Multiply $- 6$ by $- 4$ .

$a = \frac{- 2 \pm \sqrt{4 - 4 b^{2} + 24 b - 4 \cdot 10}}{2 \cdot 1}$

Multiply $- 4$ by $10$ .

$a = \frac{- 2 \pm \sqrt{4 - 4 b^{2} + 24 b - 40}}{2 \cdot 1}$

Subtract $40$ from $4$ .

$a = \frac{- 2 \pm \sqrt{- 4 b^{2} + 24 b - 36}}{2 \cdot 1}$

Rewrite $- 4 b^{2} + 24 b - 36$ in a factored form.

Factor $4$ out of $- 4 b^{2} + 24 b - 36$ .

Factor $4$ out of $- 4 b^{2}$ .

$a = \frac{- 2 \pm \sqrt{4 (- b^{2}) + 24 b - 36}}{2 \cdot 1}$

Factor $4$ out of $24 b$ .

$a = \frac{- 2 \pm \sqrt{4 (- b^{2}) + 4 (6 b) - 36}}{2 \cdot 1}$

Factor $4$ out of $- 36$ .

$a = \frac{- 2 \pm \sqrt{4 (- b^{2}) + 4 (6 b) + 4 (- 9)}}{2 \cdot 1}$

Factor $4$ out of $4 (- b^{2}) + 4 (6 b)$ .

$a = \frac{- 2 \pm \sqrt{4 (- b^{2} + 6 b) + 4 (- 9)}}{2 \cdot 1}$

Factor $4$ out of $4 (- b^{2} + 6 b) + 4 (- 9)$ .

$a = \frac{- 2 \pm \sqrt{4 (- b^{2} + 6 b - 9)}}{2 \cdot 1}$

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = - 1 \cdot - 9 = 9$ and whose sum is $b = 6$ .

Factor $6$ out of $6 b$ .

$a = \frac{- 2 \pm \sqrt{4 (- b^{2} + 6 (b) - 9)}}{2 \cdot 1}$

Rewrite $6$ as $3$ plus $3$

$a = \frac{- 2 \pm \sqrt{4 (- b^{2} + (3 + 3) b - 9)}}{2 \cdot 1}$

Apply the distributive property.

$a = \frac{- 2 \pm \sqrt{4 (- b^{2} + 3 b + 3 b - 9)}}{2 \cdot 1}$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$a = \frac{- 2 \pm \sqrt{4 ((- b^{2} + 3 b) + 3 b - 9)}}{2 \cdot 1}$

Factor out the greatest common factor (GCF) from each group.

$a = \frac{- 2 \pm \sqrt{4 (b (- b + 3) - 3 (- b + 3))}}{2 \cdot 1}$

Factor the polynomial by factoring out the greatest common factor, $- b + 3$ .

$a = \frac{- 2 \pm \sqrt{4 ((- b + 3) (b - 3))}}{2 \cdot 1}$

$a = \frac{- 2 \pm \sqrt{4 (- b + 3) (b - 3)}}{2 \cdot 1}$

Combine exponents.

Factor $- 1$ out of $- b$ .

$a = \frac{- 2 \pm \sqrt{4 (- (b) + 3) (b - 3)}}{2 \cdot 1}$

Rewrite $3$ as $- 1 (- 3)$ .

$a = \frac{- 2 \pm \sqrt{4 (- (b) - 1 \cdot - 3) (b - 3)}}{2 \cdot 1}$

Factor $- 1$ out of $- (b) - 1 (- 3)$ .

$a = \frac{- 2 \pm \sqrt{4 (- (b - 3)) (b - 3)}}{2 \cdot 1}$

Rewrite $- (b - 3)$ as $- 1 (b - 3)$ .

$a = \frac{- 2 \pm \sqrt{4 (- 1 (b - 3)) (b - 3)}}{2 \cdot 1}$

Raise $b - 3$ to the power of $1$ .

$a = \frac{- 2 \pm \sqrt{4 \cdot (- 1 ((b - 3) (b - 3)))}}{2 \cdot 1}$

Raise $b - 3$ to the power of $1$ .

$a = \frac{- 2 \pm \sqrt{4 \cdot (- 1 ((b - 3) (b - 3)))}}{2 \cdot 1}$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$a = \frac{- 2 \pm \sqrt{4 \cdot (- 1 {(b - 3)}^{1 + 1})}}{2 \cdot 1}$

Add $1$ and $1$ .

$a = \frac{- 2 \pm \sqrt{4 \cdot (- 1 {(b - 3)}^{2})}}{2 \cdot 1}$

Multiply $4$ by $- 1$ .

$a = \frac{- 2 \pm \sqrt{- 4 {(b - 3)}^{2}}}{2 \cdot 1}$

Rewrite $- 4 {(b - 3)}^{2}$ as ${(2 (b - 3))}^{2} \cdot - 1$ .

Factor $4$ out of $- 4$ .

$a = \frac{- 2 \pm \sqrt{4 (- 1) {(b - 3)}^{2}}}{2 \cdot 1}$

Rewrite $4$ as $2^{2}$ .

$a = \frac{- 2 \pm \sqrt{2^{2} \cdot (- 1 {(b - 3)}^{2})}}{2 \cdot 1}$

Move $- 1$ .

$a = \frac{- 2 \pm \sqrt{2^{2} {(b - 3)}^{2} \cdot - 1}}{2 \cdot 1}$

Rewrite $2^{2} {(b - 3)}^{2}$ as ${(2 (b - 3))}^{2}$ .

$a = \frac{- 2 \pm \sqrt{{(2 (b - 3))}^{2} \cdot - 1}}{2 \cdot 1}$

Pull terms out from under the radical.

$a = \frac{- 2 \pm 2 (b - 3) \sqrt{- 1}}{2 \cdot 1}$

Rewrite $\sqrt{- 1}$ as $i$ .

$a = \frac{- 2 \pm 2 (b - 3) i}{2 \cdot 1}$

Apply the distributive property.

$a = \frac{- 2 \pm (2 b + 2 \cdot - 3) i}{2 \cdot 1}$

Multiply $2$ by $- 3$ .

$a = \frac{- 2 \pm (2 b - 6) i}{2 \cdot 1}$

Apply the distributive property.

$a = \frac{- 2 \pm (2 b i - 6 i)}{2 \cdot 1}$

Multiply $2$ by $1$ .

$a = \frac{- 2 \pm (2 b i - 6 i)}{2}$

Step 4

The final answer is the combination of both solutions.

$a = - 1 + b i - 3 i$

$a = - 1 - b i + 3 i$

Solve for a a^2+b^2+2a-6b+10=0

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