Solve Using the Square Root Property 5t^2-2t-3=0

$5 t^{2} - 2 t - 3 = 0$

Step 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 $a \cdot c = 5 \cdot - 3 = - 15$ and whose sum is $b = - 2$ .

Factor $- 2$ out of $- 2 t$ .

$5 t^{2} - 2 t - 3 = 0$

Rewrite $- 2$ as $3$ plus $- 5$

$5 t^{2} + (3 - 5) t - 3 = 0$

Apply the distributive property.

$5 t^{2} + 3 t - 5 t - 3 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(5 t^{2} + 3 t) - 5 t - 3 = 0$

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

$t (5 t + 3) - (5 t + 3) = 0$

Factor the polynomial by factoring out the greatest common factor, $5 t + 3$ .

$(5 t + 3) (t - 1) = 0$

Step 2

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$5 t + 3 = 0$

$t - 1 = 0$

Step 3

Set $5 t + 3$ equal to $0$ and solve for $t$ .

Set $5 t + 3$ equal to $0$ .

$5 t + 3 = 0$

Solve $5 t + 3 = 0$ for $t$ .

Subtract $3$ from both sides of the equation.

$5 t = - 3$

Divide each term in $5 t = - 3$ by $5$ and simplify.

Divide each term in $5 t = - 3$ by $5$ .

$\frac{5 t}{5} = \frac{- 3}{5}$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 t}{5} = \frac{- 3}{5}$

Divide $t$ by $1$ .

$t = \frac{- 3}{5}$

Simplify the right side.

Move the negative in front of the fraction.

$t = - \frac{3}{5}$

Step 4

Set $t - 1$ equal to $0$ and solve for $t$ .

Set $t - 1$ equal to $0$ .

$t - 1 = 0$

Add $1$ to both sides of the equation.

$t = 1$

Step 5

The final solution is all the values that make $(5 t + 3) (t - 1) = 0$ true.

$t = - \frac{3}{5}, 1$

Solve Using the Square Root Property 5t^2-2t-3=0

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