Solve 1^2=0.5^2+x^2

$$1 ^ { 2 } = 0.5 ^ { 2 } + x ^ { 2 }$$

Solve for x

$$x=\frac{\sqrt{3}}{2}\approx 0.866025404$$

$$x=-\frac{\sqrt{3}}{2}\approx -0.866025404$$

Steps by Finding Square Root

Calculate 1 to the power of 2 and get 1.

$$1=0.5^{2}+x^{2}$$

Calculate 0.5 to the power of 2 and get 0.25.

$$1=0.25+x^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$0.25+x^{2}=1$$

Subtract 0.25 from both sides.

$$x^{2}=1-0.25$$

Subtract 0.25 from 1 to get 0.75.

$$x^{2}=0.75$$

Take the square root of both sides of the equation.

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

Steps Using the Quadratic Formula

Calculate 1 to the power of 2 and get 1.

$$1=0.5^{2}+x^{2}$$

Calculate 0.5 to the power of 2 and get 0.25.

$$1=0.25+x^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$0.25+x^{2}=1$$

Subtract 1 from both sides.

$$0.25+x^{2}-1=0$$

Subtract 1 from 0.25 to get -0.75.

$$-0.75+x^{2}=0$$

Quadratic equations like this one, with an x^{2} term but no x term, can still be solved using the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}, once they are put in standard form: ax^{2}+bx+c=0.

$$x^{2}-0.75=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -0.75 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-0.75\right)}}{2}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-0.75\right)}}{2}$$

Multiply -4 times -0.75.

$$x=\frac{0±\sqrt{3}}{2}$$

Now solve the equation x=\frac{0±\sqrt{3}}{2} when ± is plus.

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

Now solve the equation x=\frac{0±\sqrt{3}}{2} when ± is minus.

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

The equation is now solved.

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