Solve x^2/100+x-56=0

$$\frac { x ^ { 2 } } { 100 } + x - 56 = 0$$

Solve for x

$$x=-140$$

$$x=40$$

Steps Using Factoring

Multiply both sides of the equation by 100.

$$x^{2}+100x-5600=0$$

To solve the equation, factor x^{2}+100x-5600 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.

$$a+b=100 ab=-5600$$

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 -5600.

$$-1,5600 -2,2800 -4,1400 -5,1120 -7,800 -8,700 -10,560 -14,400 -16,350 -20,280 -25,224 -28,200 -32,175 -35,160 -40,140 -50,112 -56,100 -70,80$$

Calculate the sum for each pair.

$$-1+5600=5599 -2+2800=2798 -4+1400=1396 -5+1120=1115 -7+800=793 -8+700=692 -10+560=550 -14+400=386 -16+350=334 -20+280=260 -25+224=199 -28+200=172 -32+175=143 -35+160=125 -40+140=100 -50+112=62 -56+100=44 -70+80=10$$

The solution is the pair that gives sum 100.

$$a=-40 b=140$$

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

$$\left(x-40\right)\left(x+140\right)$$

To find equation solutions, solve x-40=0 and x+140=0.

$$x=40 x=-140$$

Steps Using Factoring By Grouping

Multiply both sides of the equation by 100.

$$x^{2}+100x-5600=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-5600. To find a and b, set up a system to be solved.

$$a+b=100 ab=1\left(-5600\right)=-5600$$

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 -5600.

$$-1,5600 -2,2800 -4,1400 -5,1120 -7,800 -8,700 -10,560 -14,400 -16,350 -20,280 -25,224 -28,200 -32,175 -35,160 -40,140 -50,112 -56,100 -70,80$$

Calculate the sum for each pair.

$$-1+5600=5599 -2+2800=2798 -4+1400=1396 -5+1120=1115 -7+800=793 -8+700=692 -10+560=550 -14+400=386 -16+350=334 -20+280=260 -25+224=199 -28+200=172 -32+175=143 -35+160=125 -40+140=100 -50+112=62 -56+100=44 -70+80=10$$

The solution is the pair that gives sum 100.

$$a=-40 b=140$$

Rewrite x^{2}+100x-5600 as \left(x^{2}-40x\right)+\left(140x-5600\right).

$$\left(x^{2}-40x\right)+\left(140x-5600\right)$$

Factor out x in the first and 140 in the second group.

$$x\left(x-40\right)+140\left(x-40\right)$$

Factor out common term x-40 by using distributive property.

$$\left(x-40\right)\left(x+140\right)$$

To find equation solutions, solve x-40=0 and x+140=0.

$$x=40 x=-140$$

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.

$$\frac{1}{100}x^{2}+x-56=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{1}{100} for a, 1 for b, and -56 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-1±\sqrt{1^{2}-4\times \frac{1}{100}\left(-56\right)}}{2\times \frac{1}{100}}$$

Square 1.

$$x=\frac{-1±\sqrt{1-4\times \frac{1}{100}\left(-56\right)}}{2\times \frac{1}{100}}$$

Multiply -4 times \frac{1}{100}.

$$x=\frac{-1±\sqrt{1-\frac{1}{25}\left(-56\right)}}{2\times \frac{1}{100}}$$

Multiply -\frac{1}{25} times -56.

$$x=\frac{-1±\sqrt{1+\frac{56}{25}}}{2\times \frac{1}{100}}$$

Add 1 to \frac{56}{25}.

$$x=\frac{-1±\sqrt{\frac{81}{25}}}{2\times \frac{1}{100}}$$

Take the square root of \frac{81}{25}.

$$x=\frac{-1±\frac{9}{5}}{2\times \frac{1}{100}}$$

Multiply 2 times \frac{1}{100}.

$$x=\frac{-1±\frac{9}{5}}{\frac{1}{50}}$$

Now solve the equation x=\frac{-1±\frac{9}{5}}{\frac{1}{50}} when ± is plus. Add -1 to \frac{9}{5}.

$$x=\frac{\frac{4}{5}}{\frac{1}{50}}$$

Divide \frac{4}{5} by \frac{1}{50} by multiplying \frac{4}{5} by the reciprocal of \frac{1}{50}.

$$x=40$$

Now solve the equation x=\frac{-1±\frac{9}{5}}{\frac{1}{50}} when ± is minus. Subtract \frac{9}{5} from -1.

$$x=-\frac{\frac{14}{5}}{\frac{1}{50}}$$

Divide -\frac{14}{5} by \frac{1}{50} by multiplying -\frac{14}{5} by the reciprocal of \frac{1}{50}.

$$x=-140$$

The equation is now solved.

$$x=40 x=-140$$

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.

$$\frac{1}{100}x^{2}+x-56=0$$

Add 56 to both sides of the equation.

$$\frac{1}{100}x^{2}+x-56-\left(-56\right)=-\left(-56\right)$$

Subtracting -56 from itself leaves 0.

$$\frac{1}{100}x^{2}+x=-\left(-56\right)$$

Subtract -56 from 0.

$$\frac{1}{100}x^{2}+x=56$$

Multiply both sides by 100.

$$\frac{\frac{1}{100}x^{2}+x}{\frac{1}{100}}=\frac{56}{\frac{1}{100}}$$

Dividing by \frac{1}{100} undoes the multiplication by \frac{1}{100}.

$$x^{2}+\frac{1}{\frac{1}{100}}x=\frac{56}{\frac{1}{100}}$$

Divide 1 by \frac{1}{100} by multiplying 1 by the reciprocal of \frac{1}{100}.

$$x^{2}+100x=\frac{56}{\frac{1}{100}}$$

Divide 56 by \frac{1}{100} by multiplying 56 by the reciprocal of \frac{1}{100}.

$$x^{2}+100x=5600$$

Divide 100, the coefficient of the x term, by 2 to get 50. Then add the square of 50 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+100x+50^{2}=5600+50^{2}$$

Square 50.

$$x^{2}+100x+2500=5600+2500$$

Add 5600 to 2500.

$$x^{2}+100x+2500=8100$$

Factor x^{2}+100x+2500. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+50\right)^{2}=8100$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+50\right)^{2}}=\sqrt{8100}$$

Simplify.

$$x+50=90 x+50=-90$$

Subtract 50 from both sides of the equation.

$$x=40 x=-140$$