Solve for x
x=-15
x=12
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-x^{2}-3x+180=0
Divide both sides by 4.
a+b=-3 ab=-180=-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 negative, the negative number has greater absolute value than the positive. 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.
-4x^{2}-12x+720=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{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\left(-4\right)\times 720}}{2\left(-4\right)}
This equation is in standard form: ax^{2}+bx+c=0. Substitute -4 for a, -12 for b, and 720 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-12\right)±\sqrt{144-4\left(-4\right)\times 720}}{2\left(-4\right)}
Square -12.
x=\frac{-\left(-12\right)±\sqrt{144+16\times 720}}{2\left(-4\right)}
Multiply -4 times -4.
x=\frac{-\left(-12\right)±\sqrt{144+11520}}{2\left(-4\right)}
Multiply 16 times 720.
x=\frac{-\left(-12\right)±\sqrt{11664}}{2\left(-4\right)}
Add 144 to 11520.
x=\frac{-\left(-12\right)±108}{2\left(-4\right)}
Take the square root of 11664.
x=\frac{12±108}{2\left(-4\right)}
The opposite of -12 is 12.
x=\frac{12±108}{-8}
Multiply 2 times -4.
x=\frac{120}{-8}
Now solve the equation x=\frac{12±108}{-8} when ± is plus. Add 12 to 108.
x=-15
Divide 120 by -8.
x=-\frac{96}{-8}
Now solve the equation x=\frac{12±108}{-8} when ± is minus. Subtract 108 from 12.
x=12
Divide -96 by -8.
x=-15 x=12
The equation is now solved.
-4x^{2}-12x+720=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.
-4x^{2}-12x+720-720=-720
Subtract 720 from both sides of the equation.
-4x^{2}-12x=-720
Subtracting 720 from itself leaves 0.
\frac{-4x^{2}-12x}{-4}=-\frac{720}{-4}
Divide both sides by -4.
x^{2}+\left(-\frac{12}{-4}\right)x=-\frac{720}{-4}
Dividing by -4 undoes the multiplication by -4.
x^{2}+3x=-\frac{720}{-4}
Divide -12 by -4.
x^{2}+3x=180
Divide -720 by -4.
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.
x ^ 2 +3x -180 = 0
Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.
r + s = -3 rs = -180
Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C
r = -\frac{3}{2} - u s = -\frac{3}{2} + u
Two numbers r and s sum up to -3 exactly when the average of the two numbers is \frac{1}{2}*-3 = -\frac{3}{2}. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>
(-\frac{3}{2} - u) (-\frac{3}{2} + u) = -180
To solve for unknown quantity u, substitute these in the product equation rs = -180
\frac{9}{4} - u^2 = -180
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -180-\frac{9}{4} = -\frac{729}{4}
Simplify the expression by subtracting \frac{9}{4} on both sides
u^2 = \frac{729}{4} u = \pm\sqrt{\frac{729}{4}} = \pm \frac{27}{2}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =-\frac{3}{2} - \frac{27}{2} = -15 s = -\frac{3}{2} + \frac{27}{2} = 12
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.
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