Solve 1.024=h^2

$$1.024 = h ^ { 2 }$$

Solve for h

$$h = \frac{8 \sqrt{10}}{25} \approx 1.011928851$$

$$h = -\frac{8 \sqrt{10}}{25} \approx -1.011928851$$

Steps by Finding Square Root

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

$$h^{2}=1.024$$

Take the square root of both sides of the equation.

$$h=\frac{8\sqrt{10}}{25} h=-\frac{8\sqrt{10}}{25}$$

Steps Using the Quadratic Formula

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

$$h^{2}=1.024$$

Subtract 1.024 from both sides.

$$h^{2}-1.024=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -1.024 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$h=\frac{0±\sqrt{0^{2}-4\left(-1.024\right)}}{2}$$

Square 0.

$$h=\frac{0±\sqrt{-4\left(-1.024\right)}}{2}$$

Multiply -4 times -1.024.

$$h=\frac{0±\sqrt{4.096}}{2}$$

Take the square root of 4.096.

$$h=\frac{0±\frac{16\sqrt{10}}{25}}{2}$$

Now solve the equation h=\frac{0±\frac{16\sqrt{10}}{25}}{2} when ± is plus.

$$h=\frac{8\sqrt{10}}{25}$$

Now solve the equation h=\frac{0±\frac{16\sqrt{10}}{25}}{2} when ± is minus.

$$h=-\frac{8\sqrt{10}}{25}$$

The equation is now solved.

$$h=\frac{8\sqrt{10}}{25} h=-\frac{8\sqrt{10}}{25}$$