Solve by Factoring -8=22t-6t^2

$- 8 = 22 t - 6 t^{2}$

Step 1

Move all the expressions to the left side of the equation.

Subtract $22 t$ from both sides of the equation.

$- 8 - 22 t = - 6 t^{2}$

Add $6 t^{2}$ to both sides of the equation.

$- 8 - 22 t + 6 t^{2} = 0$

Step 2

Factor $2$ out of $- 8 - 22 t + 6 t^{2}$ .

Factor $2$ out of $- 8$ .

$2 (- 4) - 22 t + 6 t^{2} = 0$

Factor $2$ out of $- 22 t$ .

$2 (- 4) + 2 (- 11 t) + 6 t^{2} = 0$

Factor $2$ out of $6 t^{2}$ .

$2 (- 4) + 2 (- 11 t) + 2 (3 t^{2}) = 0$

Factor $2$ out of $2 (- 4) + 2 (- 11 t)$ .

$2 (- 4 - 11 t) + 2 (3 t^{2}) = 0$

Factor $2$ out of $2 (- 4 - 11 t) + 2 (3 t^{2})$ .

$2 (- 4 - 11 t + 3 t^{2}) = 0$

Step 3

Factor.

Factor by grouping.

Reorder terms.

$2 (3 t^{2} - 11 t - 4) = 0$

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 $a \cdot c = 3 \cdot - 4 = - 12$ and whose sum is $b = - 11$ .

Factor $- 11$ out of $- 11 t$ .

$2 (3 t^{2} - 11 t - 4) = 0$

Rewrite $- 11$ as $1$ plus $- 12$

$2 (3 t^{2} + (1 - 12) t - 4) = 0$

Apply the distributive property.

$2 (3 t^{2} + 1 t - 12 t - 4) = 0$

Multiply $t$ by $1$ .

$2 (3 t^{2} + t - 12 t - 4) = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$2 ((3 t^{2} + t) - 12 t - 4) = 0$

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

$2 (t (3 t + 1) - 4 (3 t + 1)) = 0$

Factor the polynomial by factoring out the greatest common factor, $3 t + 1$ .

$2 ((3 t + 1) (t - 4)) = 0$

Remove unnecessary parentheses.

$2 (3 t + 1) (t - 4) = 0$

Step 4

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

$3 t + 1 = 0$

$t - 4 = 0$

Step 5

Set $3 t + 1$ equal to $0$ and solve for $t$ .

Set $3 t + 1$ equal to $0$ .

$3 t + 1 = 0$

Solve $3 t + 1 = 0$ for $t$ .

Subtract $1$ from both sides of the equation.

$3 t = - 1$

Divide each term in $3 t = - 1$ by $3$ and simplify.

Divide each term in $3 t = - 1$ by $3$ .

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

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

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

Divide $t$ by $1$ .

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

Simplify the right side.

Move the negative in front of the fraction.

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

Step 6

Set $t - 4$ equal to $0$ and solve for $t$ .

Set $t - 4$ equal to $0$ .

$t - 4 = 0$

Add $4$ to both sides of the equation.

$t = 4$

Step 7

The final solution is all the values that make $2 (3 t + 1) (t - 4) = 0$ true.

$t = - \frac{1}{3}, 4$

Solve by Factoring -8=22t-6t^2

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