Solve for z 18z^2-31z+12=-2z^2

$18 z^{2} - 31 z + 12 = - 2 z^{2}$

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

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

$18 z^{2} - 31 z + 12 + 2 z^{2} = 0$

Add $18 z^{2}$ and $2 z^{2}$ .

$20 z^{2} - 31 z + 12 = 0$

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 $a \cdot c = 20 \cdot 12 = 240$ and whose sum is $b = - 31$ .

Factor $- 31$ out of $- 31 z$ .

$20 z^{2} - 31 z + 12 = 0$

Rewrite $- 31$ as $- 15$ plus $- 16$

$20 z^{2} + (- 15 - 16) z + 12 = 0$

Apply the distributive property.

$20 z^{2} - 15 z - 16 z + 12 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(20 z^{2} - 15 z) - 16 z + 12 = 0$

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

$5 z (4 z - 3) - 4 (4 z - 3) = 0$

Factor the polynomial by factoring out the greatest common factor, $4 z - 3$ .

$(4 z - 3) (5 z - 4) = 0$

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

$4 z - 3 = 0$

$5 z - 4 = 0$

Set $4 z - 3$ equal to $0$ and solve for $z$ .

Set $4 z - 3$ equal to $0$ .

$4 z - 3 = 0$

Solve $4 z - 3 = 0$ for $z$ .

Add $3$ to both sides of the equation.

$4 z = 3$

Divide each term in $4 z = 3$ by $4$ and simplify.

Divide each term in $4 z = 3$ by $4$ .

$\frac{4 z}{4} = \frac{3}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 z}{4} = \frac{3}{4}$

Divide $z$ by $1$ .

$z = \frac{3}{4}$

Set $5 z - 4$ equal to $0$ and solve for $z$ .

Set $5 z - 4$ equal to $0$ .

$5 z - 4 = 0$

Solve $5 z - 4 = 0$ for $z$ .

Add $4$ to both sides of the equation.

$5 z = 4$

Divide each term in $5 z = 4$ by $5$ and simplify.

Divide each term in $5 z = 4$ by $5$ .

$\frac{5 z}{5} = \frac{4}{5}$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 z}{5} = \frac{4}{5}$

Divide $z$ by $1$ .

$z = \frac{4}{5}$

The final solution is all the values that make $(4 z - 3) (5 z - 4) = 0$ true.

$z = \frac{3}{4}, \frac{4}{5}$

The result can be shown in multiple forms.

Exact Form:

$z = \frac{3}{4}, \frac{4}{5}$

Decimal Form:

$z = 0.75, 0.8$

Solve for z 18z^2-31z+12=-2z^2

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