Solve for t 7t^2-15t+6=4

$7 t^{2} - 15 t + 6 = 4$

Step 1

Subtract $4$ from both sides of the equation.

$7 t^{2} - 15 t + 6 - 4 = 0$

Step 2

Subtract $4$ from $6$ .

$7 t^{2} - 15 t + 2 = 0$

Step 3

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 $a \cdot c = 7 \cdot 2 = 14$ and whose sum is $b = - 15$ .

Factor $- 15$ out of $- 15 t$ .

$7 t^{2} - 15 t + 2 = 0$

Rewrite $- 15$ as $- 1$ plus $- 14$

$7 t^{2} + (- 1 - 14) t + 2 = 0$

Apply the distributive property.

$7 t^{2} - 1 t - 14 t + 2 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(7 t^{2} - 1 t) - 14 t + 2 = 0$

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

$t (7 t - 1) - 2 (7 t - 1) = 0$

Factor the polynomial by factoring out the greatest common factor, $7 t - 1$ .

$(7 t - 1) (t - 2) = 0$

Step 4

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

$7 t - 1 = 0$

$t - 2 = 0$

Step 5

Set $7 t - 1$ equal to $0$ and solve for $t$ .

Set $7 t - 1$ equal to $0$ .

$7 t - 1 = 0$

Solve $7 t - 1 = 0$ for $t$ .

Add $1$ to both sides of the equation.

$7 t = 1$

Divide each term in $7 t = 1$ by $7$ and simplify.

Divide each term in $7 t = 1$ by $7$ .

$\frac{7 t}{7} = \frac{1}{7}$

Simplify the left side.

Cancel the common factor of $7$ .

Cancel the common factor.

$\frac{7 t}{7} = \frac{1}{7}$

Divide $t$ by $1$ .

$t = \frac{1}{7}$

Step 6

Set $t - 2$ equal to $0$ and solve for $t$ .

Set $t - 2$ equal to $0$ .

$t - 2 = 0$

Add $2$ to both sides of the equation.

$t = 2$

Step 7

The final solution is all the values that make $(7 t - 1) (t - 2) = 0$ true.

$t = \frac{1}{7}, 2$

Step 8

The result can be shown in multiple forms.

Exact Form:

$t = \frac{1}{7}, 2$

Decimal Form:

$t = 0.\overline{142857}, 2$

Solve for t 7t^2-15t+6=4

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