Solve 3x^2+9=36

$$3 x ^ { 2 } + 9 = 36$$

Solve for x

$$x=3$$

$$x=-3$$

Steps Using Difference of Squares

Subtract 36 from both sides.

$$3x^{2}+9-36=0$$

Subtract 36 from 9 to get -27.

$$3x^{2}-27=0$$

Divide both sides by 3.

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

Consider x^{2}-9. Rewrite x^{2}-9 as x^{2}-3^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(x-3\right)\left(x+3\right)=0$$

To find equation solutions, solve x-3=0 and x+3=0.

$$x=3 x=-3$$

Steps by Finding Square Root

Subtract 9 from both sides.

$$3x^{2}=36-9$$

Subtract 9 from 36 to get 27.

$$3x^{2}=27$$

Divide both sides by 3.

$$x^{2}=\frac{27}{3}$$

Divide 27 by 3 to get 9.

$$x^{2}=9$$

Take the square root of both sides of the equation.

$$x=3 x=-3$$

Steps Using the Quadratic Formula

Subtract 36 from both sides.

$$3x^{2}+9-36=0$$

Subtract 36 from 9 to get -27.

$$3x^{2}-27=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 0 for b, and -27 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\times 3\left(-27\right)}}{2\times 3}$$

Square 0.

$$x=\frac{0±\sqrt{-4\times 3\left(-27\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{0±\sqrt{-12\left(-27\right)}}{2\times 3}$$

Multiply -12 times -27.

$$x=\frac{0±\sqrt{324}}{2\times 3}$$

Take the square root of 324.

$$x=\frac{0±18}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{0±18}{6}$$

Now solve the equation x=\frac{0±18}{6} when ± is plus. Divide 18 by 6.

$$x=3$$

Now solve the equation x=\frac{0±18}{6} when ± is minus. Divide -18 by 6.

$$x=-3$$

The equation is now solved.

$$x=3 x=-3$$