Solve x^2+140x-1500=0

$${ x }^{ 2 } +140x-1500=0$$

Solve for x

$$x=-150$$

$$x=10$$

Steps Using Factoring

To solve the equation, factor x^{2}+140x-1500 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.

$$a+b=140 ab=-1500$$

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 -1500.

$$-1,1500 -2,750 -3,500 -4,375 -5,300 -6,250 -10,150 -12,125 -15,100 -20,75 -25,60 -30,50$$

Calculate the sum for each pair.

$$-1+1500=1499 -2+750=748 -3+500=497 -4+375=371 -5+300=295 -6+250=244 -10+150=140 -12+125=113 -15+100=85 -20+75=55 -25+60=35 -30+50=20$$

The solution is the pair that gives sum 140.

$$a=-10 b=150$$

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

$$\left(x-10\right)\left(x+150\right)$$

To find equation solutions, solve x-10=0 and x+150=0.

$$x=10 x=-150$$

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-1500. To find a and b, set up a system to be solved.

$$a+b=140 ab=1\left(-1500\right)=-1500$$

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 -1500.

$$-1,1500 -2,750 -3,500 -4,375 -5,300 -6,250 -10,150 -12,125 -15,100 -20,75 -25,60 -30,50$$

Calculate the sum for each pair.

$$-1+1500=1499 -2+750=748 -3+500=497 -4+375=371 -5+300=295 -6+250=244 -10+150=140 -12+125=113 -15+100=85 -20+75=55 -25+60=35 -30+50=20$$

The solution is the pair that gives sum 140.

$$a=-10 b=150$$

Rewrite x^{2}+140x-1500 as \left(x^{2}-10x\right)+\left(150x-1500\right).

$$\left(x^{2}-10x\right)+\left(150x-1500\right)$$

Factor out x in the first and 150 in the second group.

$$x\left(x-10\right)+150\left(x-10\right)$$

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

$$\left(x-10\right)\left(x+150\right)$$

To find equation solutions, solve x-10=0 and x+150=0.

$$x=10 x=-150$$

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.

$$x^{2}+140x-1500=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 140 for b, and -1500 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-140±\sqrt{140^{2}-4\left(-1500\right)}}{2}$$

Square 140.

$$x=\frac{-140±\sqrt{19600-4\left(-1500\right)}}{2}$$

Multiply -4 times -1500.

$$x=\frac{-140±\sqrt{19600+6000}}{2}$$

Add 19600 to 6000.

$$x=\frac{-140±\sqrt{25600}}{2}$$

Take the square root of 25600.

$$x=\frac{-140±160}{2}$$

Now solve the equation x=\frac{-140±160}{2} when ± is plus. Add -140 to 160.

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

Divide 20 by 2.

$$x=10$$

Now solve the equation x=\frac{-140±160}{2} when ± is minus. Subtract 160 from -140.

$$x=-\frac{300}{2}$$

Divide -300 by 2.

$$x=-150$$

The equation is now solved.

$$x=10 x=-150$$

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.

$$x^{2}+140x-1500=0$$

Add 1500 to both sides of the equation.

$$x^{2}+140x-1500-\left(-1500\right)=-\left(-1500\right)$$

Subtracting -1500 from itself leaves 0.

$$x^{2}+140x=-\left(-1500\right)$$

Subtract -1500 from 0.

$$x^{2}+140x=1500$$

Divide 140, the coefficient of the x term, by 2 to get 70. Then add the square of 70 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+140x+70^{2}=1500+70^{2}$$

Square 70.

$$x^{2}+140x+4900=1500+4900$$

Add 1500 to 4900.

$$x^{2}+140x+4900=6400$$

Factor x^{2}+140x+4900. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+70\right)^{2}=6400$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+70\right)^{2}}=\sqrt{6400}$$

Simplify.

$$x+70=80 x+70=-80$$

Subtract 70 from both sides of the equation.

$$x=10 x=-150$$