Solve for y 13y^2+10y-3=5y^2

$13 y^{2} + 10 y - 3 = 5 y^{2}$

Move all terms containing $y$ to the left side of the equation.

Subtract $5 y^{2}$ from both sides of the equation.

$13 y^{2} + 10 y - 3 - 5 y^{2} = 0$

Subtract $5 y^{2}$ from $13 y^{2}$ .

$8 y^{2} + 10 y - 3 = 0$

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 $a \cdot c = 8 \cdot - 3 = - 24$ and whose sum is $b = 10$ .

Factor $10$ out of $10 y$ .

$8 y^{2} + 10 (y) - 3 = 0$

Rewrite $10$ as $- 2$ plus $12$

$8 y^{2} + (- 2 + 12) y - 3 = 0$

Apply the distributive property.

$8 y^{2} - 2 y + 12 y - 3 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(8 y^{2} - 2 y) + 12 y - 3 = 0$

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

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

Factor the polynomial by factoring out the greatest common factor, $4 y - 1$ .

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

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

$4 y - 1 = 0$

$2 y + 3 = 0$

Set $4 y - 1$ equal to $0$ and solve for $y$ .

Set $4 y - 1$ equal to $0$ .

$4 y - 1 = 0$

Solve $4 y - 1 = 0$ for $y$ .

Add $1$ to both sides of the equation.

$4 y = 1$

Divide each term in $4 y = 1$ by $4$ and simplify.

Divide each term in $4 y = 1$ by $4$ .

$\frac{4 y}{4} = \frac{1}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 y}{4} = \frac{1}{4}$

Divide $y$ by $1$ .

$y = \frac{1}{4}$

Set $2 y + 3$ equal to $0$ and solve for $y$ .

Set $2 y + 3$ equal to $0$ .

$2 y + 3 = 0$

Solve $2 y + 3 = 0$ for $y$ .

Subtract $3$ from both sides of the equation.

$2 y = - 3$

Divide each term in $2 y = - 3$ by $2$ and simplify.

Divide each term in $2 y = - 3$ by $2$ .

$\frac{2 y}{2} = \frac{- 3}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 y}{2} = \frac{- 3}{2}$

Divide $y$ by $1$ .

$y = \frac{- 3}{2}$

Simplify the right side.

Move the negative in front of the fraction.

$y = - \frac{3}{2}$

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

$y = \frac{1}{4}, - \frac{3}{2}$

The result can be shown in multiple forms.

Exact Form:

$y = \frac{1}{4}, - \frac{3}{2}$

Decimal Form:

$y = 0.25, - 1.5$

Mixed Number Form:

$y = \frac{1}{4}, - 1 \frac{1}{2}$

Solve for y 13y^2+10y-3=5y^2

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