Solve V^2=25^2+(75-2V)^2

$$V ^ { 2 } = 25 ^ { 2 } + ( 75 - 2 V ) ^ { 2 }$$

Solve for V

$$V=\frac{25\sqrt{6}}{3}+50\approx 70.412414523$$

$$V=-\frac{25\sqrt{6}}{3}+50\approx 29.587585477$$

Steps Using the Quadratic Formula

Calculate 25 to the power of 2 and get 625.

$$V^{2}=625+\left(75-2V\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(75-2V\right)^{2}.

$$V^{2}=625+5625-300V+4V^{2}$$

Add 625 and 5625 to get 6250.

$$V^{2}=6250-300V+4V^{2}$$

Subtract 6250 from both sides.

$$V^{2}-6250=-300V+4V^{2}$$

Add 300V to both sides.

$$V^{2}-6250+300V=4V^{2}$$

Subtract 4V^{2} from both sides.

$$V^{2}-6250+300V-4V^{2}=0$$

Combine V^{2} and -4V^{2} to get -3V^{2}.

$$-3V^{2}-6250+300V=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-3V^{2}+300V-6250=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, 300 for b, and -6250 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$V=\frac{-300±\sqrt{300^{2}-4\left(-3\right)\left(-6250\right)}}{2\left(-3\right)}$$

Square 300.

$$V=\frac{-300±\sqrt{90000-4\left(-3\right)\left(-6250\right)}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$V=\frac{-300±\sqrt{90000+12\left(-6250\right)}}{2\left(-3\right)}$$

Multiply 12 times -6250.

$$V=\frac{-300±\sqrt{90000-75000}}{2\left(-3\right)}$$

Add 90000 to -75000.

$$V=\frac{-300±\sqrt{15000}}{2\left(-3\right)}$$

Take the square root of 15000.

$$V=\frac{-300±50\sqrt{6}}{2\left(-3\right)}$$

Multiply 2 times -3.

$$V=\frac{-300±50\sqrt{6}}{-6}$$

Now solve the equation V=\frac{-300±50\sqrt{6}}{-6} when ± is plus. Add -300 to 50\sqrt{6}.

$$V=\frac{50\sqrt{6}-300}{-6}$$

Divide -300+50\sqrt{6} by -6.

$$V=-\frac{25\sqrt{6}}{3}+50$$

Now solve the equation V=\frac{-300±50\sqrt{6}}{-6} when ± is minus. Subtract 50\sqrt{6} from -300.

$$V=\frac{-50\sqrt{6}-300}{-6}$$

Divide -300-50\sqrt{6} by -6.

$$V=\frac{25\sqrt{6}}{3}+50$$

The equation is now solved.

$$V=-\frac{25\sqrt{6}}{3}+50 V=\frac{25\sqrt{6}}{3}+50$$

Steps for Completing the Square

Calculate 25 to the power of 2 and get 625.

$$V^{2}=625+\left(75-2V\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(75-2V\right)^{2}.

$$V^{2}=625+5625-300V+4V^{2}$$

Add 625 and 5625 to get 6250.

$$V^{2}=6250-300V+4V^{2}$$

Add 300V to both sides.

$$V^{2}+300V=6250+4V^{2}$$

Subtract 4V^{2} from both sides.

$$V^{2}+300V-4V^{2}=6250$$

Combine V^{2} and -4V^{2} to get -3V^{2}.

$$-3V^{2}+300V=6250$$

Divide both sides by -3.

$$\frac{-3V^{2}+300V}{-3}=\frac{6250}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$V^{2}+\frac{300}{-3}V=\frac{6250}{-3}$$

Divide 300 by -3.

$$V^{2}-100V=\frac{6250}{-3}$$

Divide 6250 by -3.

$$V^{2}-100V=-\frac{6250}{3}$$

Divide -100, the coefficient of the x term, by 2 to get -50. Then add the square of -50 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$V^{2}-100V+\left(-50\right)^{2}=-\frac{6250}{3}+\left(-50\right)^{2}$$

Square -50.

$$V^{2}-100V+2500=-\frac{6250}{3}+2500$$

Add -\frac{6250}{3} to 2500.

$$V^{2}-100V+2500=\frac{1250}{3}$$

Factor V^{2}-100V+2500. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(V-50\right)^{2}=\frac{1250}{3}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(V-50\right)^{2}}=\sqrt{\frac{1250}{3}}$$

Simplify.

$$V-50=\frac{25\sqrt{6}}{3} V-50=-\frac{25\sqrt{6}}{3}$$

Add 50 to both sides of the equation.

$$V=\frac{25\sqrt{6}}{3}+50 V=-\frac{25\sqrt{6}}{3}+50$$