Solve 9x^2+12=21

$$9 { x }^{ 2 } +12 = 21$$

Solve for x

$$x=1$$

$$x=-1$$

Steps Using Difference of Squares

Subtract 21 from both sides.

$$9x^{2}+12-21=0$$

Subtract 21 from 12 to get -9.

$$9x^{2}-9=0$$

Divide both sides by 9.

$$x^{2}-1=0$$

Consider x^{2}-1. Rewrite x^{2}-1 as x^{2}-1^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

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

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

$$x=1 x=-1$$

Steps by Finding Square Root

Subtract 12 from both sides.

$$9x^{2}=21-12$$

Subtract 12 from 21 to get 9.

$$9x^{2}=9$$

Divide both sides by 9.

$$x^{2}=\frac{9}{9}$$

Divide 9 by 9 to get 1.

$$x^{2}=1$$

Take the square root of both sides of the equation.

$$x=1 x=-1$$

Steps Using the Quadratic Formula

Subtract 21 from both sides.

$$9x^{2}+12-21=0$$

Subtract 21 from 12 to get -9.

$$9x^{2}-9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, 0 for b, and -9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\times 9\left(-9\right)}}{2\times 9}$$

Square 0.

$$x=\frac{0±\sqrt{-4\times 9\left(-9\right)}}{2\times 9}$$

Multiply -4 times 9.

$$x=\frac{0±\sqrt{-36\left(-9\right)}}{2\times 9}$$

Multiply -36 times -9.

$$x=\frac{0±\sqrt{324}}{2\times 9}$$

Take the square root of 324.

$$x=\frac{0±18}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{0±18}{18}$$

Now solve the equation x=\frac{0±18}{18} when ± is plus. Divide 18 by 18.

$$x=1$$

Now solve the equation x=\frac{0±18}{18} when ± is minus. Divide -18 by 18.

$$x=-1$$

The equation is now solved.

$$x=1 x=-1$$