Find the Roots (Zeros) 3x^2=12-5x

$3 x^{2} = 12 - 5 x$

Add $5 x$ to both sides of the equation.

$3 x^{2} + 5 x = 12$

Subtract $12$ from both sides of the equation.

$3 x^{2} + 5 x - 12 = 0$

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 $a \cdot c = 3 \cdot - 12 = - 36$ and whose sum is $b = 5$ .

Factor $5$ out of $5 x$ .

$3 x^{2} + 5 (x) - 12 = 0$

Rewrite $5$ as $- 4$ plus $9$

$3 x^{2} + (- 4 + 9) x - 12 = 0$

Apply the distributive property.

$3 x^{2} - 4 x + 9 x - 12 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(3 x^{2} - 4 x) + 9 x - 12 = 0$

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

$x (3 x - 4) + 3 (3 x - 4) = 0$

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

$(3 x - 4) (x + 3) = 0$

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

$3 x - 4 = 0$

$x + 3 = 0$

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

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

$3 x - 4 = 0$

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

Add $4$ to both sides of the equation.

$3 x = 4$

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

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

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

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

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

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

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

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

$x = \frac{4}{3}, - 3$

Find the Roots (Zeros) 3x^2=12-5x

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