Solve V_2^2=1.44+2*2*(0.43-3/20)

$$V _ { 2 } ^ { 2 } = 1.44 + 2 \times 2 \times ( 0.43 - \frac { 3 } { 20 } )$$

Solve for V_2

$$V_{2}=1.6$$

$$V_{2}=-1.6$$

Steps Using Difference of Squares

Multiply 2 and 2 to get 4.

$$V_{2}^{2}=1.44+4\left(0.43-\frac{3}{20}\right)$$

Subtract \frac{3}{20} from 0.43 to get \frac{7}{25}.

$$V_{2}^{2}=1.44+4\times \frac{7}{25}$$

Multiply 4 and \frac{7}{25} to get \frac{28}{25}.

$$V_{2}^{2}=1.44+\frac{28}{25}$$

Add 1.44 and \frac{28}{25} to get \frac{64}{25}.

$$V_{2}^{2}=\frac{64}{25}$$

Subtract \frac{64}{25} from both sides.

$$V_{2}^{2}-\frac{64}{25}=0$$

Multiply both sides by 25.

$$25V_{2}^{2}-64=0$$

Consider 25V_{2}^{2}-64. Rewrite 25V_{2}^{2}-64 as \left(5V_{2}\right)^{2}-8^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(5V_{2}-8\right)\left(5V_{2}+8\right)=0$$

To find equation solutions, solve 5V_{2}-8=0 and 5V_{2}+8=0.

$$V_{2}=\frac{8}{5} V_{2}=-\frac{8}{5}$$

Steps by Finding Square Root

Multiply 2 and 2 to get 4.

$$V_{2}^{2}=1.44+4\left(0.43-\frac{3}{20}\right)$$

Subtract \frac{3}{20} from 0.43 to get \frac{7}{25}.

$$V_{2}^{2}=1.44+4\times \frac{7}{25}$$

Multiply 4 and \frac{7}{25} to get \frac{28}{25}.

$$V_{2}^{2}=1.44+\frac{28}{25}$$

Add 1.44 and \frac{28}{25} to get \frac{64}{25}.

$$V_{2}^{2}=\frac{64}{25}$$

Take the square root of both sides of the equation.

$$V_{2}=\frac{8}{5} V_{2}=-\frac{8}{5}$$

Steps Using the Quadratic Formula

Multiply 2 and 2 to get 4.

$$V_{2}^{2}=1.44+4\left(0.43-\frac{3}{20}\right)$$

Subtract \frac{3}{20} from 0.43 to get \frac{7}{25}.

$$V_{2}^{2}=1.44+4\times \frac{7}{25}$$

Multiply 4 and \frac{7}{25} to get \frac{28}{25}.

$$V_{2}^{2}=1.44+\frac{28}{25}$$

Add 1.44 and \frac{28}{25} to get \frac{64}{25}.

$$V_{2}^{2}=\frac{64}{25}$$

Subtract \frac{64}{25} from both sides.

$$V_{2}^{2}-\frac{64}{25}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -\frac{64}{25} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$V_{2}=\frac{0±\sqrt{0^{2}-4\left(-\frac{64}{25}\right)}}{2}$$

Square 0.

$$V_{2}=\frac{0±\sqrt{-4\left(-\frac{64}{25}\right)}}{2}$$

Multiply -4 times -\frac{64}{25}.

$$V_{2}=\frac{0±\sqrt{\frac{256}{25}}}{2}$$

Take the square root of \frac{256}{25}.

$$V_{2}=\frac{0±\frac{16}{5}}{2}$$

Now solve the equation V_{2}=\frac{0±\frac{16}{5}}{2} when ± is plus.

$$V_{2}=\frac{8}{5}$$

Now solve the equation V_{2}=\frac{0±\frac{16}{5}}{2} when ± is minus.

$$V_{2}=-\frac{8}{5}$$

The equation is now solved.

$$V_{2}=\frac{8}{5} V_{2}=-\frac{8}{5}$$