Solve 0.04+x^2=9*0.2/4

$$0.04+ { x }^{ 2 } =9 \times \frac{ 0.2 }{ 4 }$$

Solve for x

$$x=\frac{\sqrt{41}}{10}\approx 0.640312424$$

$$x=-\frac{\sqrt{41}}{10}\approx -0.640312424$$

Steps by Finding Square Root

Expand \frac{0.2}{4} by multiplying both numerator and the denominator by 10.

$$0.04+x^{2}=9\times \frac{2}{40}$$

Reduce the fraction \frac{2}{40} to lowest terms by extracting and canceling out 2.

$$0.04+x^{2}=9\times \frac{1}{20}$$

Multiply 9 and \frac{1}{20} to get \frac{9}{20}.

$$0.04+x^{2}=\frac{9}{20}$$

Subtract 0.04 from both sides.

$$x^{2}=\frac{9}{20}-0.04$$

Subtract 0.04 from \frac{9}{20} to get \frac{41}{100}.

$$x^{2}=\frac{41}{100}$$

Take the square root of both sides of the equation.

$$x=\frac{\sqrt{41}}{10} x=-\frac{\sqrt{41}}{10}$$

Steps Using the Quadratic Formula

Expand \frac{0.2}{4} by multiplying both numerator and the denominator by 10.

$$0.04+x^{2}=9\times \frac{2}{40}$$

Reduce the fraction \frac{2}{40} to lowest terms by extracting and canceling out 2.

$$0.04+x^{2}=9\times \frac{1}{20}$$

Multiply 9 and \frac{1}{20} to get \frac{9}{20}.

$$0.04+x^{2}=\frac{9}{20}$$

Subtract \frac{9}{20} from both sides.

$$0.04+x^{2}-\frac{9}{20}=0$$

Subtract \frac{9}{20} from 0.04 to get -\frac{41}{100}.

$$-\frac{41}{100}+x^{2}=0$$

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.

$$x^{2}-\frac{41}{100}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -\frac{41}{100} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-\frac{41}{100}\right)}}{2}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-\frac{41}{100}\right)}}{2}$$

Multiply -4 times -\frac{41}{100}.

$$x=\frac{0±\sqrt{\frac{41}{25}}}{2}$$

Take the square root of \frac{41}{25}.

$$x=\frac{0±\frac{\sqrt{41}}{5}}{2}$$

Now solve the equation x=\frac{0±\frac{\sqrt{41}}{5}}{2} when ± is plus.

$$x=\frac{\sqrt{41}}{10}$$

Now solve the equation x=\frac{0±\frac{\sqrt{41}}{5}}{2} when ± is minus.

$$x=-\frac{\sqrt{41}}{10}$$

The equation is now solved.

$$x=\frac{\sqrt{41}}{10} x=-\frac{\sqrt{41}}{10}$$