Solve 9x^2+12x-6=x^2-7

$$9 x ^ { 2 } + 12 x - 6 = x ^ { 2 } - 7$$

Solve for x

$$x=\frac{\sqrt{7}-3}{4}\approx -0.088562172$$

$$x=\frac{-\sqrt{7}-3}{4}\approx -1.411437828$$

Steps Using the Quadratic Formula

Subtract x^{2} from both sides.

$$9x^{2}+12x-6-x^{2}=-7$$

Combine 9x^{2} and -x^{2} to get 8x^{2}.

$$8x^{2}+12x-6=-7$$

Add 7 to both sides.

$$8x^{2}+12x-6+7=0$$

Add -6 and 7 to get 1.

$$8x^{2}+12x+1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 8 for a, 12 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-12±\sqrt{12^{2}-4\times 8}}{2\times 8}$$

Square 12.

$$x=\frac{-12±\sqrt{144-4\times 8}}{2\times 8}$$

Multiply -4 times 8.

$$x=\frac{-12±\sqrt{144-32}}{2\times 8}$$

Add 144 to -32.

$$x=\frac{-12±\sqrt{112}}{2\times 8}$$

Take the square root of 112.

$$x=\frac{-12±4\sqrt{7}}{2\times 8}$$

Multiply 2 times 8.

$$x=\frac{-12±4\sqrt{7}}{16}$$

Now solve the equation x=\frac{-12±4\sqrt{7}}{16} when ± is plus. Add -12 to 4\sqrt{7}.

$$x=\frac{4\sqrt{7}-12}{16}$$

Divide -12+4\sqrt{7} by 16.

$$x=\frac{\sqrt{7}-3}{4}$$

Now solve the equation x=\frac{-12±4\sqrt{7}}{16} when ± is minus. Subtract 4\sqrt{7} from -12.

$$x=\frac{-4\sqrt{7}-12}{16}$$

Divide -12-4\sqrt{7} by 16.

$$x=\frac{-\sqrt{7}-3}{4}$$

The equation is now solved.

$$x=\frac{\sqrt{7}-3}{4} x=\frac{-\sqrt{7}-3}{4}$$

Steps for Completing the Square

Subtract x^{2} from both sides.

$$9x^{2}+12x-6-x^{2}=-7$$

Combine 9x^{2} and -x^{2} to get 8x^{2}.

$$8x^{2}+12x-6=-7$$

Add 6 to both sides.

$$8x^{2}+12x=-7+6$$

Add -7 and 6 to get -1.

$$8x^{2}+12x=-1$$

Divide both sides by 8.

$$\frac{8x^{2}+12x}{8}=-\frac{1}{8}$$

Dividing by 8 undoes the multiplication by 8.

$$x^{2}+\frac{12}{8}x=-\frac{1}{8}$$

Reduce the fraction \frac{12}{8} to lowest terms by extracting and canceling out 4.

$$x^{2}+\frac{3}{2}x=-\frac{1}{8}$$

Divide \frac{3}{2}, the coefficient of the x term, by 2 to get \frac{3}{4}. Then add the square of \frac{3}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{3}{2}x+\left(\frac{3}{4}\right)^{2}=-\frac{1}{8}+\left(\frac{3}{4}\right)^{2}$$

Square \frac{3}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{3}{2}x+\frac{9}{16}=-\frac{1}{8}+\frac{9}{16}$$

Add -\frac{1}{8} to \frac{9}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{3}{2}x+\frac{9}{16}=\frac{7}{16}$$

Factor x^{2}+\frac{3}{2}x+\frac{9}{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}.

$$\left(x+\frac{3}{4}\right)^{2}=\frac{7}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{4}\right)^{2}}=\sqrt{\frac{7}{16}}$$

Simplify.

$$x+\frac{3}{4}=\frac{\sqrt{7}}{4} x+\frac{3}{4}=-\frac{\sqrt{7}}{4}$$

Subtract \frac{3}{4} from both sides of the equation.

$$x=\frac{\sqrt{7}-3}{4} x=\frac{-\sqrt{7}-3}{4}$$