Solve 0.28=1^2+2^2-x^2/2*1*2

$$0.28 = \frac { 1 ^ { 2 } + 2 ^ { 2 } - x ^ { 2 } } { 2 \times 1 \times 2 }$$

Solve for x

$$x = \frac{\sqrt{97}}{5} \approx 1.96977156$$

$$x = -\frac{\sqrt{97}}{5} \approx -1.96977156$$

Steps by Finding Square Root

Calculate 1 to the power of 2 and get 1.

$$0.28=\frac{1+2^{2}-x^{2}}{2\times 1\times 2}$$

Calculate 2 to the power of 2 and get 4.

$$0.28=\frac{1+4-x^{2}}{2\times 1\times 2}$$

Add 1 and 4 to get 5.

$$0.28=\frac{5-x^{2}}{2\times 1\times 2}$$

Multiply 2 and 1 to get 2.

$$0.28=\frac{5-x^{2}}{2\times 2}$$

Multiply 2 and 2 to get 4.

$$0.28=\frac{5-x^{2}}{4}$$

Divide each term of 5-x^{2} by 4 to get \frac{5}{4}-\frac{1}{4}x^{2}.

$$0.28=\frac{5}{4}-\frac{1}{4}x^{2}$$

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

$$\frac{5}{4}-\frac{1}{4}x^{2}=0.28$$

Subtract \frac{5}{4} from both sides.

$$-\frac{1}{4}x^{2}=0.28-\frac{5}{4}$$

Subtract \frac{5}{4} from 0.28 to get -\frac{97}{100}.

$$-\frac{1}{4}x^{2}=-\frac{97}{100}$$

Multiply both sides by -4, the reciprocal of -\frac{1}{4}.

$$x^{2}=-\frac{97}{100}\left(-4\right)$$

Multiply -\frac{97}{100} and -4 to get \frac{97}{25}.

$$x^{2}=\frac{97}{25}$$

Take the square root of both sides of the equation.

$$x=\frac{\sqrt{97}}{5} x=-\frac{\sqrt{97}}{5}$$

Steps Using the Quadratic Formula

Calculate 1 to the power of 2 and get 1.

$$0.28=\frac{1+2^{2}-x^{2}}{2\times 1\times 2}$$

Calculate 2 to the power of 2 and get 4.

$$0.28=\frac{1+4-x^{2}}{2\times 1\times 2}$$

Add 1 and 4 to get 5.

$$0.28=\frac{5-x^{2}}{2\times 1\times 2}$$

Multiply 2 and 1 to get 2.

$$0.28=\frac{5-x^{2}}{2\times 2}$$

Multiply 2 and 2 to get 4.

$$0.28=\frac{5-x^{2}}{4}$$

Divide each term of 5-x^{2} by 4 to get \frac{5}{4}-\frac{1}{4}x^{2}.

$$0.28=\frac{5}{4}-\frac{1}{4}x^{2}$$

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

$$\frac{5}{4}-\frac{1}{4}x^{2}=0.28$$

Subtract 0.28 from both sides.

$$\frac{5}{4}-\frac{1}{4}x^{2}-0.28=0$$

Subtract 0.28 from \frac{5}{4} to get \frac{97}{100}.

$$\frac{97}{100}-\frac{1}{4}x^{2}=0$$

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.

$$-\frac{1}{4}x^{2}+\frac{97}{100}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -\frac{1}{4} for a, 0 for b, and \frac{97}{100} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-\frac{1}{4}\right)\times \frac{97}{100}}}{2\left(-\frac{1}{4}\right)}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-\frac{1}{4}\right)\times \frac{97}{100}}}{2\left(-\frac{1}{4}\right)}$$

Multiply -4 times -\frac{1}{4}.

$$x=\frac{0±\sqrt{\frac{97}{100}}}{2\left(-\frac{1}{4}\right)}$$

Take the square root of \frac{97}{100}.

$$x=\frac{0±\frac{\sqrt{97}}{10}}{2\left(-\frac{1}{4}\right)}$$

Multiply 2 times -\frac{1}{4}.

$$x=\frac{0±\frac{\sqrt{97}}{10}}{-\frac{1}{2}}$$

Now solve the equation x=\frac{0±\frac{\sqrt{97}}{10}}{-\frac{1}{2}} when ± is plus.

$$x=-\frac{\sqrt{97}}{5}$$

Now solve the equation x=\frac{0±\frac{\sqrt{97}}{10}}{-\frac{1}{2}} when ± is minus.

$$x=\frac{\sqrt{97}}{5}$$

The equation is now solved.

$$x=-\frac{\sqrt{97}}{5} x=\frac{\sqrt{97}}{5}$$