Solve y^2/2-25y-200=0

$$\frac { y ^ { 2 } } { 2 } - 25 y - 200 = 0$$

Solve for y

$$y=5\sqrt{41}+25\approx 57.015621187$$

$$y=25-5\sqrt{41}\approx -7.015621187$$

Steps Using the Quadratic Formula

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.

$$\frac{1}{2}y^{2}-25y-200=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{1}{2} for a, -25 for b, and -200 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-25\right)±\sqrt{\left(-25\right)^{2}-4\times \frac{1}{2}\left(-200\right)}}{2\times \frac{1}{2}}$$

Square -25.

$$y=\frac{-\left(-25\right)±\sqrt{625-4\times \frac{1}{2}\left(-200\right)}}{2\times \frac{1}{2}}$$

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

$$y=\frac{-\left(-25\right)±\sqrt{625-2\left(-200\right)}}{2\times \frac{1}{2}}$$

Multiply -2 times -200.

$$y=\frac{-\left(-25\right)±\sqrt{625+400}}{2\times \frac{1}{2}}$$

Add 625 to 400.

$$y=\frac{-\left(-25\right)±\sqrt{1025}}{2\times \frac{1}{2}}$$

Take the square root of 1025.

$$y=\frac{-\left(-25\right)±5\sqrt{41}}{2\times \frac{1}{2}}$$

The opposite of -25 is 25.

$$y=\frac{25±5\sqrt{41}}{2\times \frac{1}{2}}$$

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

$$y=\frac{25±5\sqrt{41}}{1}$$

Now solve the equation y=\frac{25±5\sqrt{41}}{1} when ± is plus. Add 25 to 5\sqrt{41}.

$$y=\frac{5\sqrt{41}+25}{1}$$

Divide 25+5\sqrt{41} by 1.

$$y=5\sqrt{41}+25$$

Now solve the equation y=\frac{25±5\sqrt{41}}{1} when ± is minus. Subtract 5\sqrt{41} from 25.

$$y=\frac{25-5\sqrt{41}}{1}$$

Divide 25-5\sqrt{41} by 1.

$$y=25-5\sqrt{41}$$

The equation is now solved.

$$y=5\sqrt{41}+25 y=25-5\sqrt{41}$$

Steps for Completing the Square

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$\frac{1}{2}y^{2}-25y-200=0$$

Add 200 to both sides of the equation.

$$\frac{1}{2}y^{2}-25y-200-\left(-200\right)=-\left(-200\right)$$

Subtracting -200 from itself leaves 0.

$$\frac{1}{2}y^{2}-25y=-\left(-200\right)$$

Subtract -200 from 0.

$$\frac{1}{2}y^{2}-25y=200$$

Multiply both sides by 2.

$$\frac{\frac{1}{2}y^{2}-25y}{\frac{1}{2}}=\frac{200}{\frac{1}{2}}$$

Dividing by \frac{1}{2} undoes the multiplication by \frac{1}{2}.

$$y^{2}+\left(-\frac{25}{\frac{1}{2}}\right)y=\frac{200}{\frac{1}{2}}$$

Divide -25 by \frac{1}{2} by multiplying -25 by the reciprocal of \frac{1}{2}.

$$y^{2}-50y=\frac{200}{\frac{1}{2}}$$

Divide 200 by \frac{1}{2} by multiplying 200 by the reciprocal of \frac{1}{2}.

$$y^{2}-50y=400$$

Divide -50, the coefficient of the x term, by 2 to get -25. Then add the square of -25 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}-50y+\left(-25\right)^{2}=400+\left(-25\right)^{2}$$

Square -25.

$$y^{2}-50y+625=400+625$$

Add 400 to 625.

$$y^{2}-50y+625=1025$$

Factor y^{2}-50y+625. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(y-25\right)^{2}=1025$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y-25\right)^{2}}=\sqrt{1025}$$

Simplify.

$$y-25=5\sqrt{41} y-25=-5\sqrt{41}$$

Add 25 to both sides of the equation.

$$y=5\sqrt{41}+25 y=25-5\sqrt{41}$$