Solve 3x^2-4=8

$$3 x ^ { 2 } - 4 = 8$$

Solve for x

$$x=2$$

$$x=-2$$

Steps Using Difference of Squares

Subtract 8 from both sides.

$$3x^{2}-4-8=0$$

Subtract 8 from -4 to get -12.

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

Divide both sides by 3.

$$x^{2}-4=0$$

Consider x^{2}-4. Rewrite x^{2}-4 as x^{2}-2^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

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

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

$$x=2 x=-2$$

Steps by Finding Square Root

Add 4 to both sides.

$$3x^{2}=8+4$$

Add 8 and 4 to get 12.

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

Divide both sides by 3.

$$x^{2}=\frac{12}{3}$$

Divide 12 by 3 to get 4.

$$x^{2}=4$$

Take the square root of both sides of the equation.

$$x=2 x=-2$$

Steps Using the Quadratic Formula

Subtract 8 from both sides.

$$3x^{2}-4-8=0$$

Subtract 8 from -4 to get -12.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 0 for b, and -12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\times 3\left(-12\right)}}{2\times 3}$$

Square 0.

$$x=\frac{0±\sqrt{-4\times 3\left(-12\right)}}{2\times 3}$$

Multiply -4 times 3.

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

Multiply -12 times -12.

$$x=\frac{0±\sqrt{144}}{2\times 3}$$

Take the square root of 144.

$$x=\frac{0±12}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{0±12}{6}$$

Now solve the equation x=\frac{0±12}{6} when ± is plus. Divide 12 by 6.

$$x=2$$

Now solve the equation x=\frac{0±12}{6} when ± is minus. Divide -12 by 6.

$$x=-2$$

The equation is now solved.

$$x=2 x=-2$$