Solve 3x^2-6=(x+2)(x-3)

$$3 { x }^{ 2 } -6=(x+2)(x-3)$$

Solve for x

$$x=-\frac{1}{2}=-0.5$$

$$x=0$$

Steps Using Factoring

Use the distributive property to multiply x+2 by x-3 and combine like terms.

$$3x^{2}-6=x^{2}-x-6$$

Subtract x^{2} from both sides.

$$3x^{2}-6-x^{2}=-x-6$$

Combine 3x^{2} and -x^{2} to get 2x^{2}.

$$2x^{2}-6=-x-6$$

Add x to both sides.

$$2x^{2}-6+x=-6$$

Add 6 to both sides.

$$2x^{2}-6+x+6=0$$

Add -6 and 6 to get 0.

$$2x^{2}+x=0$$

Factor out x.

$$x\left(2x+1\right)=0$$

To find equation solutions, solve x=0 and 2x+1=0.

$$x=0 x=-\frac{1}{2}$$

Steps Using the Quadratic Formula

Use the distributive property to multiply x+2 by x-3 and combine like terms.

$$3x^{2}-6=x^{2}-x-6$$

Subtract x^{2} from both sides.

$$3x^{2}-6-x^{2}=-x-6$$

Combine 3x^{2} and -x^{2} to get 2x^{2}.

$$2x^{2}-6=-x-6$$

Add x to both sides.

$$2x^{2}-6+x=-6$$

Add 6 to both sides.

$$2x^{2}-6+x+6=0$$

Add -6 and 6 to get 0.

$$2x^{2}+x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 1 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-1±\sqrt{1^{2}}}{2\times 2}$$

Take the square root of 1^{2}.

$$x=\frac{-1±1}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-1±1}{4}$$

Now solve the equation x=\frac{-1±1}{4} when ± is plus. Add -1 to 1.

$$x=\frac{0}{4}$$

Divide 0 by 4.

$$x=0$$

Now solve the equation x=\frac{-1±1}{4} when ± is minus. Subtract 1 from -1.

$$x=-\frac{2}{4}$$

Reduce the fraction \frac{-2}{4} to lowest terms by extracting and canceling out 2.

$$x=-\frac{1}{2}$$

The equation is now solved.

$$x=0 x=-\frac{1}{2}$$

Steps for Completing the Square

Use the distributive property to multiply x+2 by x-3 and combine like terms.

$$3x^{2}-6=x^{2}-x-6$$

Subtract x^{2} from both sides.

$$3x^{2}-6-x^{2}=-x-6$$

Combine 3x^{2} and -x^{2} to get 2x^{2}.

$$2x^{2}-6=-x-6$$

Add x to both sides.

$$2x^{2}-6+x=-6$$

Add 6 to both sides.

$$2x^{2}+x=-6+6$$

Add -6 and 6 to get 0.

$$2x^{2}+x=0$$

Divide both sides by 2.

$$\frac{2x^{2}+x}{2}=\frac{0}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{1}{2}x=\frac{0}{2}$$

Divide 0 by 2.

$$x^{2}+\frac{1}{2}x=0$$

Divide \frac{1}{2}, the coefficient of the x term, by 2 to get \frac{1}{4}. Then add the square of \frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{1}{2}x+\left(\frac{1}{4}\right)^{2}=\left(\frac{1}{4}\right)^{2}$$

Square \frac{1}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{1}{16}$$

Factor x^{2}+\frac{1}{2}x+\frac{1}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{4}\right)^{2}=\frac{1}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{4}\right)^{2}}=\sqrt{\frac{1}{16}}$$

Simplify.

$$x+\frac{1}{4}=\frac{1}{4} x+\frac{1}{4}=-\frac{1}{4}$$

Subtract \frac{1}{4} from both sides of the equation.

$$x=0 x=-\frac{1}{2}$$