Solve 32x+20x-x^2=100

$$32x+20x- { x }^{ 2 } = 100$$

Solve for x

$$x=2$$

$$x=50$$

Steps Using Factoring By Grouping

Combine 32x and 20x to get 52x.

$$52x-x^{2}=100$$

Subtract 100 from both sides.

$$52x-x^{2}-100=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-x^{2}+52x-100=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-100. To find a and b, set up a system to be solved.

$$a+b=52 ab=-\left(-100\right)=100$$

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 100.

$$1,100 2,50 4,25 5,20 10,10$$

Calculate the sum for each pair.

$$1+100=101 2+50=52 4+25=29 5+20=25 10+10=20$$

The solution is the pair that gives sum 52.

$$a=50 b=2$$

Rewrite -x^{2}+52x-100 as \left(-x^{2}+50x\right)+\left(2x-100\right).

$$\left(-x^{2}+50x\right)+\left(2x-100\right)$$

Factor out -x in the first and 2 in the second group.

$$-x\left(x-50\right)+2\left(x-50\right)$$

Factor out common term x-50 by using distributive property.

$$\left(x-50\right)\left(-x+2\right)$$

To find equation solutions, solve x-50=0 and -x+2=0.

$$x=50 x=2$$

Steps Using the Quadratic Formula

Combine 32x and 20x to get 52x.

$$52x-x^{2}=100$$

Subtract 100 from both sides.

$$52x-x^{2}-100=0$$

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.

$$-x^{2}+52x-100=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 52 for b, and -100 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-52±\sqrt{52^{2}-4\left(-1\right)\left(-100\right)}}{2\left(-1\right)}$$

Square 52.

$$x=\frac{-52±\sqrt{2704-4\left(-1\right)\left(-100\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-52±\sqrt{2704+4\left(-100\right)}}{2\left(-1\right)}$$

Multiply 4 times -100.

$$x=\frac{-52±\sqrt{2704-400}}{2\left(-1\right)}$$

Add 2704 to -400.

$$x=\frac{-52±\sqrt{2304}}{2\left(-1\right)}$$

Take the square root of 2304.

$$x=\frac{-52±48}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-52±48}{-2}$$

Now solve the equation x=\frac{-52±48}{-2} when ± is plus. Add -52 to 48.

$$x=-\frac{4}{-2}$$

Divide -4 by -2.

$$x=2$$

Now solve the equation x=\frac{-52±48}{-2} when ± is minus. Subtract 48 from -52.

$$x=-\frac{100}{-2}$$

Divide -100 by -2.

$$x=50$$

The equation is now solved.

$$x=2 x=50$$

Steps for Completing the Square

Combine 32x and 20x to get 52x.

$$52x-x^{2}=100$$

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.

$$-x^{2}+52x=100$$

Divide both sides by -1.

$$\frac{-x^{2}+52x}{-1}=\frac{100}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{52}{-1}x=\frac{100}{-1}$$

Divide 52 by -1.

$$x^{2}-52x=\frac{100}{-1}$$

Divide 100 by -1.

$$x^{2}-52x=-100$$

Divide -52, the coefficient of the x term, by 2 to get -26. Then add the square of -26 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-52x+\left(-26\right)^{2}=-100+\left(-26\right)^{2}$$

Square -26.

$$x^{2}-52x+676=-100+676$$

Add -100 to 676.

$$x^{2}-52x+676=576$$

Factor x^{2}-52x+676. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-26\right)^{2}=576$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-26\right)^{2}}=\sqrt{576}$$

Simplify.

$$x-26=24 x-26=-24$$

Add 26 to both sides of the equation.

$$x=50 x=2$$