Solve left(2xright)^2+1=19

$${ \left(2x \right) }^{ 2 } +1=19$$

Solve for x

$$x = \frac{3 \sqrt{2}}{2} \approx 2.121320344$$

$$x = -\frac{3 \sqrt{2}}{2} \approx -2.121320344$$

Steps by Finding Square Root

Expand \left(2x\right)^{2}.

$$2^{2}x^{2}+1=19$$

Calculate 2 to the power of 2 and get 4.

$$4x^{2}+1=19$$

Subtract 1 from both sides.

$$4x^{2}=19-1$$

Subtract 1 from 19 to get 18.

$$4x^{2}=18$$

Divide both sides by 4.

$$x^{2}=\frac{18}{4}$$

Reduce the fraction \frac{18}{4} to lowest terms by extracting and canceling out 2.

$$x^{2}=\frac{9}{2}$$

Take the square root of both sides of the equation.

$$x=\frac{3\sqrt{2}}{2} x=-\frac{3\sqrt{2}}{2}$$