Solve by Factoring 3y+2y^2=20

$3 y + 2 y^{2} = 20$

Step 1

Subtract $20$ from both sides of the equation.

$3 y + 2 y^{2} - 20 = 0$

Step 2

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

$3 u + 2 u^{2} - 20 = 0$

Step 3

Factor by grouping.

Reorder terms.

$2 u^{2} + 3 u - 20 = 0$

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 $a \cdot c = 2 \cdot - 20 = - 40$ and whose sum is $b = 3$ .

Factor $3$ out of $3 u$ .

$2 u^{2} + 3 (u) - 20 = 0$

Rewrite $3$ as $- 5$ plus $8$

$2 u^{2} + (- 5 + 8) u - 20 = 0$

Apply the distributive property.

$2 u^{2} - 5 u + 8 u - 20 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(2 u^{2} - 5 u) + 8 u - 20 = 0$

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

$u (2 u - 5) + 4 (2 u - 5) = 0$

Factor the polynomial by factoring out the greatest common factor, $2 u - 5$ .

$(2 u - 5) (u + 4) = 0$

Step 4

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

$(2 y - 5) (y + 4) = 0$

Step 5

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

$2 y - 5 = 0$

$y + 4 = 0$

Step 6

Set $2 y - 5$ equal to $0$ and solve for $y$ .

Set $2 y - 5$ equal to $0$ .

$2 y - 5 = 0$

Solve $2 y - 5 = 0$ for $y$ .

Add $5$ to both sides of the equation.

$2 y = 5$

Divide each term in $2 y = 5$ by $2$ and simplify.

Divide each term in $2 y = 5$ by $2$ .

$\frac{2 y}{2} = \frac{5}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 y}{2} = \frac{5}{2}$

Divide $y$ by $1$ .

$y = \frac{5}{2}$

Step 7

Set $y + 4$ equal to $0$ and solve for $y$ .

Set $y + 4$ equal to $0$ .

$y + 4 = 0$

Subtract $4$ from both sides of the equation.

$y = - 4$

Step 8

The final solution is all the values that make $(2 y - 5) (y + 4) = 0$ true.

$y = \frac{5}{2}, - 4$

Solve by Factoring 3y+2y^2=20

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