Solve (2r)^2=0.3^2+0.2

$$( 2 r ) ^ { 2 } = 0.3 ^ { 2 } + 0.2$$

Solve for r

$$r=\frac{\sqrt{29}}{20}\approx 0.26925824$$

$$r=-\frac{\sqrt{29}}{20}\approx -0.26925824$$

Steps by Finding Square Root

Expand \left(2r\right)^{2}.

$$2^{2}r^{2}=0.3^{2}+0.2$$

Calculate 2 to the power of 2 and get 4.

$$4r^{2}=0.3^{2}+0.2$$

Calculate 0.3 to the power of 2 and get 0.09.

$$4r^{2}=0.09+0.2$$

Add 0.09 and 0.2 to get 0.29.

$$4r^{2}=0.29$$

Divide both sides by 4.

$$r^{2}=\frac{0.29}{4}$$

Expand \frac{0.29}{4} by multiplying both numerator and the denominator by 100.

$$r^{2}=\frac{29}{400}$$

Take the square root of both sides of the equation.

$$r=\frac{\sqrt{29}}{20} r=-\frac{\sqrt{29}}{20}$$

Steps Using the Quadratic Formula

Expand \left(2r\right)^{2}.

$$2^{2}r^{2}=0.3^{2}+0.2$$

Calculate 2 to the power of 2 and get 4.

$$4r^{2}=0.3^{2}+0.2$$

Calculate 0.3 to the power of 2 and get 0.09.

$$4r^{2}=0.09+0.2$$

Add 0.09 and 0.2 to get 0.29.

$$4r^{2}=0.29$$

Subtract 0.29 from both sides.

$$4r^{2}-0.29=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, 0 for b, and -0.29 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$r=\frac{0±\sqrt{0^{2}-4\times 4\left(-0.29\right)}}{2\times 4}$$

Square 0.

$$r=\frac{0±\sqrt{-4\times 4\left(-0.29\right)}}{2\times 4}$$

Multiply -4 times 4.

$$r=\frac{0±\sqrt{-16\left(-0.29\right)}}{2\times 4}$$

Multiply -16 times -0.29.

$$r=\frac{0±\sqrt{4.64}}{2\times 4}$$

Take the square root of 4.64.

$$r=\frac{0±\frac{2\sqrt{29}}{5}}{2\times 4}$$

Multiply 2 times 4.

$$r=\frac{0±\frac{2\sqrt{29}}{5}}{8}$$

Now solve the equation r=\frac{0±\frac{2\sqrt{29}}{5}}{8} when ± is plus.

$$r=\frac{\sqrt{29}}{20}$$

Now solve the equation r=\frac{0±\frac{2\sqrt{29}}{5}}{8} when ± is minus.

$$r=-\frac{\sqrt{29}}{20}$$

The equation is now solved.

$$r=\frac{\sqrt{29}}{20} r=-\frac{\sqrt{29}}{20}$$