Solve 2x^2-33=39

$$2 x ^ { 2 } - 33 = 39$$

Solve for x

$$x=6$$

$$x=-6$$

Steps Using Difference of Squares

Subtract 39 from both sides.

$$2x^{2}-33-39=0$$

Subtract 39 from -33 to get -72.

$$2x^{2}-72=0$$

Divide both sides by 2.

$$x^{2}-36=0$$

Consider x^{2}-36. Rewrite x^{2}-36 as x^{2}-6^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(x-6\right)\left(x+6\right)=0$$

To find equation solutions, solve x-6=0 and x+6=0.

$$x=6 x=-6$$

Steps by Finding Square Root

Add 33 to both sides.

$$2x^{2}=39+33$$

Add 39 and 33 to get 72.

$$2x^{2}=72$$

Divide both sides by 2.

$$x^{2}=\frac{72}{2}$$

Divide 72 by 2 to get 36.

$$x^{2}=36$$

Take the square root of both sides of the equation.

$$x=6 x=-6$$

Steps Using the Quadratic Formula

Subtract 39 from both sides.

$$2x^{2}-33-39=0$$

Subtract 39 from -33 to get -72.

$$2x^{2}-72=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 0 for b, and -72 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\times 2\left(-72\right)}}{2\times 2}$$

Square 0.

$$x=\frac{0±\sqrt{-4\times 2\left(-72\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{0±\sqrt{-8\left(-72\right)}}{2\times 2}$$

Multiply -8 times -72.

$$x=\frac{0±\sqrt{576}}{2\times 2}$$

Take the square root of 576.

$$x=\frac{0±24}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{0±24}{4}$$

Now solve the equation x=\frac{0±24}{4} when ± is plus. Divide 24 by 4.

$$x=6$$

Now solve the equation x=\frac{0±24}{4} when ± is minus. Divide -24 by 4.

$$x=-6$$

The equation is now solved.

$$x=6 x=-6$$