Solve for x
x=-30
x=15
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a+b=15 ab=-450
To solve the equation, factor x^{2}+15x-450 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.
-1,450 -2,225 -3,150 -5,90 -6,75 -9,50 -10,45 -15,30 -18,25
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 -450.
-1+450=449 -2+225=223 -3+150=147 -5+90=85 -6+75=69 -9+50=41 -10+45=35 -15+30=15 -18+25=7
Calculate the sum for each pair.
a=-15 b=30
The solution is the pair that gives sum 15.
\left(x-15\right)\left(x+30\right)
Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.
x=15 x=-30
To find equation solutions, solve x-15=0 and x+30=0.
a+b=15 ab=1\left(-450\right)=-450
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-450. To find a and b, set up a system to be solved.
-1,450 -2,225 -3,150 -5,90 -6,75 -9,50 -10,45 -15,30 -18,25
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 -450.
-1+450=449 -2+225=223 -3+150=147 -5+90=85 -6+75=69 -9+50=41 -10+45=35 -15+30=15 -18+25=7
Calculate the sum for each pair.
a=-15 b=30
The solution is the pair that gives sum 15.
\left(x^{2}-15x\right)+\left(30x-450\right)
Rewrite x^{2}+15x-450 as \left(x^{2}-15x\right)+\left(30x-450\right).
x\left(x-15\right)+30\left(x-15\right)
Factor out x in the first and 30 in the second group.
\left(x-15\right)\left(x+30\right)
Factor out common term x-15 by using distributive property.
x=15 x=-30
To find equation solutions, solve x-15=0 and x+30=0.
x^{2}+15x-450=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{-15±\sqrt{15^{2}-4\left(-450\right)}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 15 for b, and -450 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-15±\sqrt{225-4\left(-450\right)}}{2}
Square 15.
x=\frac{-15±\sqrt{225+1800}}{2}
Multiply -4 times -450.
x=\frac{-15±\sqrt{2025}}{2}
Add 225 to 1800.
x=\frac{-15±45}{2}
Take the square root of 2025.
x=\frac{30}{2}
Now solve the equation x=\frac{-15±45}{2} when ± is plus. Add -15 to 45.
x=15
Divide 30 by 2.
x=-\frac{60}{2}
Now solve the equation x=\frac{-15±45}{2} when ± is minus. Subtract 45 from -15.
x=-30
Divide -60 by 2.
x=15 x=-30
The equation is now solved.
x^{2}+15x-450=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}+15x-450-\left(-450\right)=-\left(-450\right)
Add 450 to both sides of the equation.
x^{2}+15x=-\left(-450\right)
Subtracting -450 from itself leaves 0.
x^{2}+15x=450
Subtract -450 from 0.
x^{2}+15x+\left(\frac{15}{2}\right)^{2}=450+\left(\frac{15}{2}\right)^{2}
Divide 15, the coefficient of the x term, by 2 to get \frac{15}{2}. Then add the square of \frac{15}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+15x+\frac{225}{4}=450+\frac{225}{4}
Square \frac{15}{2} by squaring both the numerator and the denominator of the fraction.
x^{2}+15x+\frac{225}{4}=\frac{2025}{4}
Add 450 to \frac{225}{4}.
\left(x+\frac{15}{2}\right)^{2}=\frac{2025}{4}
Factor x^{2}+15x+\frac{225}{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{15}{2}\right)^{2}}=\sqrt{\frac{2025}{4}}
Take the square root of both sides of the equation.
x+\frac{15}{2}=\frac{45}{2} x+\frac{15}{2}=-\frac{45}{2}
Simplify.
x=15 x=-30
Subtract \frac{15}{2} from both sides of the equation.
x ^ 2 +15x -450 = 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 = -15 rs = -450
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{15}{2} - u s = -\frac{15}{2} + u
Two numbers r and s sum up to -15 exactly when the average of the two numbers is \frac{1}{2}*-15 = -\frac{15}{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{15}{2} - u) (-\frac{15}{2} + u) = -450
To solve for unknown quantity u, substitute these in the product equation rs = -450
\frac{225}{4} - u^2 = -450
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -450-\frac{225}{4} = -\frac{2025}{4}
Simplify the expression by subtracting \frac{225}{4} on both sides
u^2 = \frac{2025}{4} u = \pm\sqrt{\frac{2025}{4}} = \pm \frac{45}{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{15}{2} - \frac{45}{2} = -30 s = -\frac{15}{2} + \frac{45}{2} = 15
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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