Solve 2x^2-3=15

$$2 x ^ { 2 } - 3 = 15$$

Solve for x

$$x=3$$

$$x=-3$$

Steps Using Difference of Squares

Subtract 15 from both sides.

$$2x^{2}-3-15=0$$

Subtract 15 from -3 to get -18.

$$2x^{2}-18=0$$

Divide both sides by 2.

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

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

Add 3 to both sides.

$$2x^{2}=15+3$$

Add 15 and 3 to get 18.

$$2x^{2}=18$$

Divide both sides by 2.

$$x^{2}=\frac{18}{2}$$

Divide 18 by 2 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 15 from both sides.

$$2x^{2}-3-15=0$$

Subtract 15 from -3 to get -18.

$$2x^{2}-18=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 0 for b, and -18 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\times 2\left(-18\right)}}{2\times 2}$$

Square 0.

$$x=\frac{0±\sqrt{-4\times 2\left(-18\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{0±\sqrt{-8\left(-18\right)}}{2\times 2}$$

Multiply -8 times -18.

$$x=\frac{0±\sqrt{144}}{2\times 2}$$

Take the square root of 144.

$$x=\frac{0±12}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{0±12}{4}$$

Now solve the equation x=\frac{0±12}{4} when ± is plus. Divide 12 by 4.

$$x=3$$

Now solve the equation x=\frac{0±12}{4} when ± is minus. Divide -12 by 4.

$$x=-3$$

The equation is now solved.

$$x=3 x=-3$$