Solve 180x-360=98x+2x^2

$$180x-360=98x+2 { x }^{ 2 }$$

Solve for x

$$x=5$$

$$x=36$$

Steps Using Factoring By Grouping

Subtract 98x from both sides.

$$180x-360-98x=2x^{2}$$

Combine 180x and -98x to get 82x.

$$82x-360=2x^{2}$$

Subtract 2x^{2} from both sides.

$$82x-360-2x^{2}=0$$

Divide both sides by 2.

$$41x-180-x^{2}=0$$

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

$$-x^{2}+41x-180=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-180. To find a and b, set up a system to be solved.

$$a+b=41 ab=-\left(-180\right)=180$$

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 180.

$$1,180 2,90 3,60 4,45 5,36 6,30 9,20 10,18 12,15$$

Calculate the sum for each pair.

$$1+180=181 2+90=92 3+60=63 4+45=49 5+36=41 6+30=36 9+20=29 10+18=28 12+15=27$$

The solution is the pair that gives sum 41.

$$a=36 b=5$$

Rewrite -x^{2}+41x-180 as \left(-x^{2}+36x\right)+\left(5x-180\right).

$$\left(-x^{2}+36x\right)+\left(5x-180\right)$$

Factor out -x in the first and 5 in the second group.

$$-x\left(x-36\right)+5\left(x-36\right)$$

Factor out common term x-36 by using distributive property.

$$\left(x-36\right)\left(-x+5\right)$$

To find equation solutions, solve x-36=0 and -x+5=0.

$$x=36 x=5$$

Steps Using the Quadratic Formula

Subtract 98x from both sides.

$$180x-360-98x=2x^{2}$$

Combine 180x and -98x to get 82x.

$$82x-360=2x^{2}$$

Subtract 2x^{2} from both sides.

$$82x-360-2x^{2}=0$$

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.

$$-2x^{2}+82x-360=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, 82 for b, and -360 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-82±\sqrt{82^{2}-4\left(-2\right)\left(-360\right)}}{2\left(-2\right)}$$

Square 82.

$$x=\frac{-82±\sqrt{6724-4\left(-2\right)\left(-360\right)}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-82±\sqrt{6724+8\left(-360\right)}}{2\left(-2\right)}$$

Multiply 8 times -360.

$$x=\frac{-82±\sqrt{6724-2880}}{2\left(-2\right)}$$

Add 6724 to -2880.

$$x=\frac{-82±\sqrt{3844}}{2\left(-2\right)}$$

Take the square root of 3844.

$$x=\frac{-82±62}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{-82±62}{-4}$$

Now solve the equation x=\frac{-82±62}{-4} when ± is plus. Add -82 to 62.

$$x=-\frac{20}{-4}$$

Divide -20 by -4.

$$x=5$$

Now solve the equation x=\frac{-82±62}{-4} when ± is minus. Subtract 62 from -82.

$$x=-\frac{144}{-4}$$

Divide -144 by -4.

$$x=36$$

The equation is now solved.

$$x=5 x=36$$

Steps for Completing the Square

Subtract 98x from both sides.

$$180x-360-98x=2x^{2}$$

Combine 180x and -98x to get 82x.

$$82x-360=2x^{2}$$

Subtract 2x^{2} from both sides.

$$82x-360-2x^{2}=0$$

Add 360 to both sides. Anything plus zero gives itself.

$$82x-2x^{2}=360$$

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.

$$-2x^{2}+82x=360$$

Divide both sides by -2.

$$\frac{-2x^{2}+82x}{-2}=\frac{360}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$x^{2}+\frac{82}{-2}x=\frac{360}{-2}$$

Divide 82 by -2.

$$x^{2}-41x=\frac{360}{-2}$$

Divide 360 by -2.

$$x^{2}-41x=-180$$

Divide -41, the coefficient of the x term, by 2 to get -\frac{41}{2}. Then add the square of -\frac{41}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-41x+\left(-\frac{41}{2}\right)^{2}=-180+\left(-\frac{41}{2}\right)^{2}$$

Square -\frac{41}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-41x+\frac{1681}{4}=-180+\frac{1681}{4}$$

Add -180 to \frac{1681}{4}.

$$x^{2}-41x+\frac{1681}{4}=\frac{961}{4}$$

Factor x^{2}-41x+\frac{1681}{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}.

$$\left(x-\frac{41}{2}\right)^{2}=\frac{961}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{41}{2}\right)^{2}}=\sqrt{\frac{961}{4}}$$

Simplify.

$$x-\frac{41}{2}=\frac{31}{2} x-\frac{41}{2}=-\frac{31}{2}$$

Add \frac{41}{2} to both sides of the equation.

$$x=36 x=5$$