Solve 2x^2+3=13

$$2 x ^ { 2 } + 3 = 13$$

Solve for x

$$x=\sqrt{5}\approx 2.236067977$$

$$x=-\sqrt{5}\approx -2.236067977$$

Steps by Finding Square Root

Subtract 3 from both sides.

$$2x^{2}=13-3$$

Subtract 3 from 13 to get 10.

$$2x^{2}=10$$

Divide both sides by 2.

$$x^{2}=\frac{10}{2}$$

Divide 10 by 2 to get 5.

$$x^{2}=5$$

Take the square root of both sides of the equation.

$$x=\sqrt{5} x=-\sqrt{5}$$

Steps Using the Quadratic Formula

Subtract 13 from both sides.

$$2x^{2}+3-13=0$$

Subtract 13 from 3 to get -10.

$$2x^{2}-10=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 0 for b, and -10 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\times 2\left(-10\right)}}{2\times 2}$$

Square 0.

$$x=\frac{0±\sqrt{-4\times 2\left(-10\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{0±\sqrt{-8\left(-10\right)}}{2\times 2}$$

Multiply -8 times -10.

$$x=\frac{0±\sqrt{80}}{2\times 2}$$

Take the square root of 80.

$$x=\frac{0±4\sqrt{5}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{0±4\sqrt{5}}{4}$$

Now solve the equation x=\frac{0±4\sqrt{5}}{4} when ± is plus.

$$x=\sqrt{5}$$

Now solve the equation x=\frac{0±4\sqrt{5}}{4} when ± is minus.

$$x=-\sqrt{5}$$

The equation is now solved.

$$x=\sqrt{5} x=-\sqrt{5}$$