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$\exponential{(x)}{2} + 3 x - 180 = 0 $
Solve for x
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a+b=3 ab=-180
To solve the equation, factor x^{2}+3x-180 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.
-1,180 -2,90 -3,60 -4,45 -5,36 -6,30 -9,20 -10,18 -12,15
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 -180.
-1+180=179 -2+90=88 -3+60=57 -4+45=41 -5+36=31 -6+30=24 -9+20=11 -10+18=8 -12+15=3
Calculate the sum for each pair.
a=-12 b=15
The solution is the pair that gives sum 3.
\left(x-12\right)\left(x+15\right)
Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.
x=12 x=-15
To find equation solutions, solve x-12=0 and x+15=0.
a+b=3 ab=1\left(-180\right)=-180
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-180. To find a and b, set up a system to be solved.
-1,180 -2,90 -3,60 -4,45 -5,36 -6,30 -9,20 -10,18 -12,15
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 -180.
-1+180=179 -2+90=88 -3+60=57 -4+45=41 -5+36=31 -6+30=24 -9+20=11 -10+18=8 -12+15=3
Calculate the sum for each pair.
a=-12 b=15
The solution is the pair that gives sum 3.
\left(x^{2}-12x\right)+\left(15x-180\right)
Rewrite x^{2}+3x-180 as \left(x^{2}-12x\right)+\left(15x-180\right).
x\left(x-12\right)+15\left(x-12\right)
Factor out x in the first and 15 in the second group.
\left(x-12\right)\left(x+15\right)
Factor out common term x-12 by using distributive property.
x=12 x=-15
To find equation solutions, solve x-12=0 and x+15=0.
x^{2}+3x-180=0
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=\frac{-3±\sqrt{3^{2}-4\left(-180\right)}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 3 for b, and -180 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-3±\sqrt{9-4\left(-180\right)}}{2}
Square 3.
x=\frac{-3±\sqrt{9+720}}{2}
Multiply -4 times -180.
x=\frac{-3±\sqrt{729}}{2}
Add 9 to 720.
x=\frac{-3±27}{2}
Take the square root of 729.
x=\frac{24}{2}
Now solve the equation x=\frac{-3±27}{2} when ± is plus. Add -3 to 27.
x=12
Divide 24 by 2.
x=-\frac{30}{2}
Now solve the equation x=\frac{-3±27}{2} when ± is minus. Subtract 27 from -3.
x=-15
Divide -30 by 2.
x=12 x=-15
The equation is now solved.
x^{2}+3x-180=0
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}+3x-180-\left(-180\right)=-\left(-180\right)
Add 180 to both sides of the equation.
x^{2}+3x=-\left(-180\right)
Subtracting -180 from itself leaves 0.
x^{2}+3x=180
Subtract -180 from 0.
x^{2}+3x+\left(\frac{3}{2}\right)^{2}=180+\left(\frac{3}{2}\right)^{2}
Divide 3, the coefficient of the x term, by 2 to get \frac{3}{2}. Then add the square of \frac{3}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+3x+\frac{9}{4}=180+\frac{9}{4}
Square \frac{3}{2} by squaring both the numerator and the denominator of the fraction.
x^{2}+3x+\frac{9}{4}=\frac{729}{4}
Add 180 to \frac{9}{4}.
\left(x+\frac{3}{2}\right)^{2}=\frac{729}{4}
Factor x^{2}+3x+\frac{9}{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{3}{2}\right)^{2}}=\sqrt{\frac{729}{4}}
Take the square root of both sides of the equation.
x+\frac{3}{2}=\frac{27}{2} x+\frac{3}{2}=-\frac{27}{2}
Simplify.
x=12 x=-15
Subtract \frac{3}{2} from both sides of the equation.