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4x^{2}+4x-120=0
Subtract 120 from both sides.
x^{2}+x-30=0
Divide both sides by 4.
a+b=1 ab=1\left(-30\right)=-30
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-30. To find a and b, set up a system to be solved.
-1,30 -2,15 -3,10 -5,6
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 -30.
-1+30=29 -2+15=13 -3+10=7 -5+6=1
Calculate the sum for each pair.
a=-5 b=6
The solution is the pair that gives sum 1.
\left(x^{2}-5x\right)+\left(6x-30\right)
Rewrite x^{2}+x-30 as \left(x^{2}-5x\right)+\left(6x-30\right).
x\left(x-5\right)+6\left(x-5\right)
Factor out x in the first and 6 in the second group.
\left(x-5\right)\left(x+6\right)
Factor out common term x-5 by using distributive property.
x=5 x=-6
To find equation solutions, solve x-5=0 and x+6=0.
4x^{2}+4x=120
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.
4x^{2}+4x-120=120-120
Subtract 120 from both sides of the equation.
4x^{2}+4x-120=0
Subtracting 120 from itself leaves 0.
x=\frac{-4±\sqrt{4^{2}-4\times 4\left(-120\right)}}{2\times 4}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, 4 for b, and -120 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-4±\sqrt{16-4\times 4\left(-120\right)}}{2\times 4}
Square 4.
x=\frac{-4±\sqrt{16-16\left(-120\right)}}{2\times 4}
Multiply -4 times 4.
x=\frac{-4±\sqrt{16+1920}}{2\times 4}
Multiply -16 times -120.
x=\frac{-4±\sqrt{1936}}{2\times 4}
Add 16 to 1920.
x=\frac{-4±44}{2\times 4}
Take the square root of 1936.
x=\frac{-4±44}{8}
Multiply 2 times 4.
x=\frac{40}{8}
Now solve the equation x=\frac{-4±44}{8} when ± is plus. Add -4 to 44.
x=5
Divide 40 by 8.
x=-\frac{48}{8}
Now solve the equation x=\frac{-4±44}{8} when ± is minus. Subtract 44 from -4.
x=-6
Divide -48 by 8.
x=5 x=-6
The equation is now solved.
4x^{2}+4x=120
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{4x^{2}+4x}{4}=\frac{120}{4}
Divide both sides by 4.
x^{2}+\frac{4}{4}x=\frac{120}{4}
Dividing by 4 undoes the multiplication by 4.
x^{2}+x=\frac{120}{4}
Divide 4 by 4.
x^{2}+x=30
Divide 120 by 4.
x^{2}+x+\left(\frac{1}{2}\right)^{2}=30+\left(\frac{1}{2}\right)^{2}
Divide 1, the coefficient of the x term, by 2 to get \frac{1}{2}. Then add the square of \frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+x+\frac{1}{4}=30+\frac{1}{4}
Square \frac{1}{2} by squaring both the numerator and the denominator of the fraction.
x^{2}+x+\frac{1}{4}=\frac{121}{4}
Add 30 to \frac{1}{4}.
\left(x+\frac{1}{2}\right)^{2}=\frac{121}{4}
Factor x^{2}+x+\frac{1}{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}.
\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{\frac{121}{4}}
Take the square root of both sides of the equation.
x+\frac{1}{2}=\frac{11}{2} x+\frac{1}{2}=-\frac{11}{2}
Simplify.
x=5 x=-6
Subtract \frac{1}{2} from both sides of the equation.