Solve 1.2^2=0.4^2+x^2

$${ 1.2 }^{ 2 } = { 0.4 }^{ 2 } + { x }^{ 2 }$$

Solve for x

$$x = \frac{4 \sqrt{2}}{5} \approx 1.13137085$$

$$x = -\frac{4 \sqrt{2}}{5} \approx -1.13137085$$

Steps by Finding Square Root

Calculate 1.2 to the power of 2 and get 1.44.

$$1.44=0.4^{2}+x^{2}$$

Calculate 0.4 to the power of 2 and get 0.16.

$$1.44=0.16+x^{2}$$

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

$$0.16+x^{2}=1.44$$

Subtract 0.16 from both sides.

$$x^{2}=1.44-0.16$$

Subtract 0.16 from 1.44 to get 1.28.

$$x^{2}=1.28$$

Take the square root of both sides of the equation.

$$x=\frac{4\sqrt{2}}{5} x=-\frac{4\sqrt{2}}{5}$$

Steps Using the Quadratic Formula

Calculate 1.2 to the power of 2 and get 1.44.

$$1.44=0.4^{2}+x^{2}$$

Calculate 0.4 to the power of 2 and get 0.16.

$$1.44=0.16+x^{2}$$

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

$$0.16+x^{2}=1.44$$

Subtract 1.44 from both sides.

$$0.16+x^{2}-1.44=0$$

Subtract 1.44 from 0.16 to get -1.28.

$$-1.28+x^{2}=0$$

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.

$$x^{2}-1.28=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -1.28 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-1.28\right)}}{2}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-1.28\right)}}{2}$$

Multiply -4 times -1.28.

$$x=\frac{0±\sqrt{5.12}}{2}$$

Take the square root of 5.12.

$$x=\frac{0±\frac{8\sqrt{2}}{5}}{2}$$

Now solve the equation x=\frac{0±\frac{8\sqrt{2}}{5}}{2} when ± is plus.

$$x=\frac{4\sqrt{2}}{5}$$

Now solve the equation x=\frac{0±\frac{8\sqrt{2}}{5}}{2} when ± is minus.

$$x=-\frac{4\sqrt{2}}{5}$$

The equation is now solved.

$$x=\frac{4\sqrt{2}}{5} x=-\frac{4\sqrt{2}}{5}$$