Solve -y^2+55y-750=0

$$- { y }^{ 2 } +55y-750=0$$

Solve for y

$$y=25$$

$$y=30$$

Steps Using Factoring By Grouping

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -y^{2}+ay+by-750. To find a and b, set up a system to be solved.

$$a+b=55 ab=-\left(-750\right)=750$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 750.

$$1,750 2,375 3,250 5,150 6,125 10,75 15,50 25,30$$

Calculate the sum for each pair.

$$1+750=751 2+375=377 3+250=253 5+150=155 6+125=131 10+75=85 15+50=65 25+30=55$$

The solution is the pair that gives sum 55.

$$a=30 b=25$$

Rewrite -y^{2}+55y-750 as \left(-y^{2}+30y\right)+\left(25y-750\right).

$$\left(-y^{2}+30y\right)+\left(25y-750\right)$$

Factor out -y in the first and 25 in the second group.

$$-y\left(y-30\right)+25\left(y-30\right)$$

Factor out common term y-30 by using distributive property.

$$\left(y-30\right)\left(-y+25\right)$$

To find equation solutions, solve y-30=0 and -y+25=0.

$$y=30 y=25$$

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.

$$-y^{2}+55y-750=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 55 for b, and -750 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-55±\sqrt{55^{2}-4\left(-1\right)\left(-750\right)}}{2\left(-1\right)}$$

Square 55.

$$y=\frac{-55±\sqrt{3025-4\left(-1\right)\left(-750\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$y=\frac{-55±\sqrt{3025+4\left(-750\right)}}{2\left(-1\right)}$$

Multiply 4 times -750.

$$y=\frac{-55±\sqrt{3025-3000}}{2\left(-1\right)}$$

Add 3025 to -3000.

$$y=\frac{-55±\sqrt{25}}{2\left(-1\right)}$$

Take the square root of 25.

$$y=\frac{-55±5}{2\left(-1\right)}$$

Multiply 2 times -1.

$$y=\frac{-55±5}{-2}$$

Now solve the equation y=\frac{-55±5}{-2} when ± is plus. Add -55 to 5.

$$y=-\frac{50}{-2}$$

Divide -50 by -2.

$$y=25$$

Now solve the equation y=\frac{-55±5}{-2} when ± is minus. Subtract 5 from -55.

$$y=-\frac{60}{-2}$$

Divide -60 by -2.

$$y=30$$

The equation is now solved.

$$y=25 y=30$$

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.

$$-y^{2}+55y-750=0$$

Add 750 to both sides of the equation.

$$-y^{2}+55y-750-\left(-750\right)=-\left(-750\right)$$

Subtracting -750 from itself leaves 0.

$$-y^{2}+55y=-\left(-750\right)$$

Subtract -750 from 0.

$$-y^{2}+55y=750$$

Divide both sides by -1.

$$\frac{-y^{2}+55y}{-1}=\frac{750}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$y^{2}+\frac{55}{-1}y=\frac{750}{-1}$$

Divide 55 by -1.

$$y^{2}-55y=\frac{750}{-1}$$

Divide 750 by -1.

$$y^{2}-55y=-750$$

Divide -55, the coefficient of the x term, by 2 to get -\frac{55}{2}. Then add the square of -\frac{55}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}-55y+\left(-\frac{55}{2}\right)^{2}=-750+\left(-\frac{55}{2}\right)^{2}$$

Square -\frac{55}{2} by squaring both the numerator and the denominator of the fraction.

$$y^{2}-55y+\frac{3025}{4}=-750+\frac{3025}{4}$$

Add -750 to \frac{3025}{4}.

$$y^{2}-55y+\frac{3025}{4}=\frac{25}{4}$$

Factor y^{2}-55y+\frac{3025}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(y-\frac{55}{2}\right)^{2}=\frac{25}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y-\frac{55}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Simplify.

$$y-\frac{55}{2}=\frac{5}{2} y-\frac{55}{2}=-\frac{5}{2}$$

Add \frac{55}{2} to both sides of the equation.

$$y=30 y=25$$