Solve 2x^2=19

$$2 x ^ { 2 } = 19$$

Solve for x

$$x = \frac{\sqrt{38}}{2} \approx 3.082207001$$

$$x = -\frac{\sqrt{38}}{2} \approx -3.082207001$$

Steps by Finding Square Root

Divide both sides by 2.

$$x^{2}=\frac{19}{2}$$

Take the square root of both sides of the equation.

$$x=\frac{\sqrt{38}}{2} x=-\frac{\sqrt{38}}{2}$$

Steps Using the Quadratic Formula

Divide both sides by 2.

$$x^{2}=\frac{19}{2}$$

Subtract \frac{19}{2} from both sides.

$$x^{2}-\frac{19}{2}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -\frac{19}{2} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-\frac{19}{2}\right)}}{2}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-\frac{19}{2}\right)}}{2}$$

Multiply -4 times -\frac{19}{2}.

$$x=\frac{0±\sqrt{38}}{2}$$

Now solve the equation x=\frac{0±\sqrt{38}}{2} when ± is plus.

$$x=\frac{\sqrt{38}}{2}$$

Now solve the equation x=\frac{0±\sqrt{38}}{2} when ± is minus.

$$x=-\frac{\sqrt{38}}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{38}}{2} x=-\frac{\sqrt{38}}{2}$$