Solve by Factoring 17y+7y^2=12

$17 y + 7 y^{2} = 12$

Step 1

Subtract $12$ from both sides of the equation.

$17 y + 7 y^{2} - 12 = 0$

Step 2

Let $u = y$ . Substitute $u$ for all occurrences of $y$ .

$17 u + 7 u^{2} - 12 = 0$

Step 3

Factor by grouping.

Reorder terms.

$7 u^{2} + 17 u - 12 = 0$

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 $a \cdot c = 7 \cdot - 12 = - 84$ and whose sum is $b = 17$ .

Factor $17$ out of $17 u$ .

$7 u^{2} + 17 (u) - 12 = 0$

Rewrite $17$ as $- 4$ plus $21$

$7 u^{2} + (- 4 + 21) u - 12 = 0$

Apply the distributive property.

$7 u^{2} - 4 u + 21 u - 12 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(7 u^{2} - 4 u) + 21 u - 12 = 0$

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

$u (7 u - 4) + 3 (7 u - 4) = 0$

Factor the polynomial by factoring out the greatest common factor, $7 u - 4$ .

$(7 u - 4) (u + 3) = 0$

Step 4

Replace all occurrences of $u$ with $y$ .

$(7 y - 4) (y + 3) = 0$

Step 5

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

$7 y - 4 = 0$

$y + 3 = 0$

Step 6

Set $7 y - 4$ equal to $0$ and solve for $y$ .

Set $7 y - 4$ equal to $0$ .

$7 y - 4 = 0$

Solve $7 y - 4 = 0$ for $y$ .

Add $4$ to both sides of the equation.

$7 y = 4$

Divide each term in $7 y = 4$ by $7$ and simplify.

Divide each term in $7 y = 4$ by $7$ .

$\frac{7 y}{7} = \frac{4}{7}$

Simplify the left side.

Cancel the common factor of $7$ .

Cancel the common factor.

$\frac{7 y}{7} = \frac{4}{7}$

Divide $y$ by $1$ .

$y = \frac{4}{7}$

Step 7

Set $y + 3$ equal to $0$ and solve for $y$ .

Set $y + 3$ equal to $0$ .

$y + 3 = 0$

Subtract $3$ from both sides of the equation.

$y = - 3$

Step 8

The final solution is all the values that make $(7 y - 4) (y + 3) = 0$ true.

$y = \frac{4}{7}, - 3$

Solve by Factoring 17y+7y^2=12

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