Solve 120x^2+8000x-92000=0

$$120 { x }^{ 2 } +8000x-92000=0$$

Solve for x

$$x = -\frac{230}{3} = -76\frac{2}{3} \approx -76.666666667$$

$$x=10$$

Steps Using Factoring By Grouping

Divide both sides by 40.

$$3x^{2}+200x-2300=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3x^{2}+ax+bx-2300. To find a and b, set up a system to be solved.

$$a+b=200 ab=3\left(-2300\right)=-6900$$

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 -6900.

$$-1,6900 -2,3450 -3,2300 -4,1725 -5,1380 -6,1150 -10,690 -12,575 -15,460 -20,345 -23,300 -25,276 -30,230 -46,150 -50,138 -60,115 -69,100 -75,92$$

Calculate the sum for each pair.

$$-1+6900=6899 -2+3450=3448 -3+2300=2297 -4+1725=1721 -5+1380=1375 -6+1150=1144 -10+690=680 -12+575=563 -15+460=445 -20+345=325 -23+300=277 -25+276=251 -30+230=200 -46+150=104 -50+138=88 -60+115=55 -69+100=31 -75+92=17$$

The solution is the pair that gives sum 200.

$$a=-30 b=230$$

Rewrite 3x^{2}+200x-2300 as \left(3x^{2}-30x\right)+\left(230x-2300\right).

$$\left(3x^{2}-30x\right)+\left(230x-2300\right)$$

Factor out 3x in the first and 230 in the second group.

$$3x\left(x-10\right)+230\left(x-10\right)$$

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

$$\left(x-10\right)\left(3x+230\right)$$

To find equation solutions, solve x-10=0 and 3x+230=0.

$$x=10 x=-\frac{230}{3}$$

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.

$$120x^{2}+8000x-92000=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 120 for a, 8000 for b, and -92000 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-8000±\sqrt{8000^{2}-4\times 120\left(-92000\right)}}{2\times 120}$$

Square 8000.

$$x=\frac{-8000±\sqrt{64000000-4\times 120\left(-92000\right)}}{2\times 120}$$

Multiply -4 times 120.

$$x=\frac{-8000±\sqrt{64000000-480\left(-92000\right)}}{2\times 120}$$

Multiply -480 times -92000.

$$x=\frac{-8000±\sqrt{64000000+44160000}}{2\times 120}$$

Add 64000000 to 44160000.

$$x=\frac{-8000±\sqrt{108160000}}{2\times 120}$$

Take the square root of 108160000.

$$x=\frac{-8000±10400}{2\times 120}$$

Multiply 2 times 120.

$$x=\frac{-8000±10400}{240}$$

Now solve the equation x=\frac{-8000±10400}{240} when ± is plus. Add -8000 to 10400.

$$x=\frac{2400}{240}$$

Divide 2400 by 240.

$$x=10$$

Now solve the equation x=\frac{-8000±10400}{240} when ± is minus. Subtract 10400 from -8000.

$$x=-\frac{18400}{240}$$

Reduce the fraction \frac{-18400}{240} to lowest terms by extracting and canceling out 80.

$$x=-\frac{230}{3}$$

The equation is now solved.

$$x=10 x=-\frac{230}{3}$$

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.

$$120x^{2}+8000x-92000=0$$

Add 92000 to both sides of the equation.

$$120x^{2}+8000x-92000-\left(-92000\right)=-\left(-92000\right)$$

Subtracting -92000 from itself leaves 0.

$$120x^{2}+8000x=-\left(-92000\right)$$

Subtract -92000 from 0.

$$120x^{2}+8000x=92000$$

Divide both sides by 120.

$$\frac{120x^{2}+8000x}{120}=\frac{92000}{120}$$

Dividing by 120 undoes the multiplication by 120.

$$x^{2}+\frac{8000}{120}x=\frac{92000}{120}$$

Reduce the fraction \frac{8000}{120} to lowest terms by extracting and canceling out 40.

$$x^{2}+\frac{200}{3}x=\frac{92000}{120}$$

Reduce the fraction \frac{92000}{120} to lowest terms by extracting and canceling out 40.

$$x^{2}+\frac{200}{3}x=\frac{2300}{3}$$

Divide \frac{200}{3}, the coefficient of the x term, by 2 to get \frac{100}{3}. Then add the square of \frac{100}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{200}{3}x+\left(\frac{100}{3}\right)^{2}=\frac{2300}{3}+\left(\frac{100}{3}\right)^{2}$$

Square \frac{100}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{200}{3}x+\frac{10000}{9}=\frac{2300}{3}+\frac{10000}{9}$$

Add \frac{2300}{3} to \frac{10000}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{200}{3}x+\frac{10000}{9}=\frac{16900}{9}$$

Factor x^{2}+\frac{200}{3}x+\frac{10000}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{100}{3}\right)^{2}=\frac{16900}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{100}{3}\right)^{2}}=\sqrt{\frac{16900}{9}}$$

Simplify.

$$x+\frac{100}{3}=\frac{130}{3} x+\frac{100}{3}=-\frac{130}{3}$$

Subtract \frac{100}{3} from both sides of the equation.

$$x=10 x=-\frac{230}{3}$$