Solve x^2+198x-2080=0

$${ x }^{ 2 } +198x-2080 = 0$$

Solve for x

$$x=-208$$

$$x=10$$

Steps Using Factoring

To solve the equation, factor x^{2}+198x-2080 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=198 ab=-2080$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -2080.

$$-1,2080 -2,1040 -4,520 -5,416 -8,260 -10,208 -13,160 -16,130 -20,104 -26,80 -32,65 -40,52$$

Calculate the sum for each pair.

$$-1+2080=2079 -2+1040=1038 -4+520=516 -5+416=411 -8+260=252 -10+208=198 -13+160=147 -16+130=114 -20+104=84 -26+80=54 -32+65=33 -40+52=12$$

The solution is the pair that gives sum 198.

$$a=-10 b=208$$

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

$$\left(x-10\right)\left(x+208\right)$$

To find equation solutions, solve x-10=0 and x+208=0.

$$x=10 x=-208$$

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 x^{2}+ax+bx-2080. To find a and b, set up a system to be solved.

$$a+b=198 ab=1\left(-2080\right)=-2080$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -2080.

$$-1,2080 -2,1040 -4,520 -5,416 -8,260 -10,208 -13,160 -16,130 -20,104 -26,80 -32,65 -40,52$$

Calculate the sum for each pair.

$$-1+2080=2079 -2+1040=1038 -4+520=516 -5+416=411 -8+260=252 -10+208=198 -13+160=147 -16+130=114 -20+104=84 -26+80=54 -32+65=33 -40+52=12$$

The solution is the pair that gives sum 198.

$$a=-10 b=208$$

Rewrite x^{2}+198x-2080 as \left(x^{2}-10x\right)+\left(208x-2080\right).

$$\left(x^{2}-10x\right)+\left(208x-2080\right)$$

Factor out x in the first and 208 in the second group.

$$x\left(x-10\right)+208\left(x-10\right)$$

Factor out common term x-10 by using distributive property.

$$\left(x-10\right)\left(x+208\right)$$

To find equation solutions, solve x-10=0 and x+208=0.

$$x=10 x=-208$$

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.

$$x^{2}+198x-2080=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 198 for b, and -2080 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-198±\sqrt{198^{2}-4\left(-2080\right)}}{2}$$

Square 198.

$$x=\frac{-198±\sqrt{39204-4\left(-2080\right)}}{2}$$

Multiply -4 times -2080.

$$x=\frac{-198±\sqrt{39204+8320}}{2}$$

Add 39204 to 8320.

$$x=\frac{-198±\sqrt{47524}}{2}$$

Take the square root of 47524.

$$x=\frac{-198±218}{2}$$

Now solve the equation x=\frac{-198±218}{2} when ± is plus. Add -198 to 218.

$$x=\frac{20}{2}$$

Divide 20 by 2.

$$x=10$$

Now solve the equation x=\frac{-198±218}{2} when ± is minus. Subtract 218 from -198.

$$x=-\frac{416}{2}$$

Divide -416 by 2.

$$x=-208$$

The equation is now solved.

$$x=10 x=-208$$

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.

$$x^{2}+198x-2080=0$$

Add 2080 to both sides of the equation.

$$x^{2}+198x-2080-\left(-2080\right)=-\left(-2080\right)$$

Subtracting -2080 from itself leaves 0.

$$x^{2}+198x=-\left(-2080\right)$$

Subtract -2080 from 0.

$$x^{2}+198x=2080$$

Divide 198, the coefficient of the x term, by 2 to get 99. Then add the square of 99 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+198x+99^{2}=2080+99^{2}$$

Square 99.

$$x^{2}+198x+9801=2080+9801$$

Add 2080 to 9801.

$$x^{2}+198x+9801=11881$$

Factor x^{2}+198x+9801. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+99\right)^{2}=11881$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+99\right)^{2}}=\sqrt{11881}$$

Simplify.

$$x+99=109 x+99=-109$$

Subtract 99 from both sides of the equation.

$$x=10 x=-208$$