Solve for x
x=-5
x=12
Graph
Share
Copied to clipboard
x^{2}-7x-60=0
Divide both sides by 8.
a+b=-7 ab=1\left(-60\right)=-60
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-60. To find a and b, set up a system to be solved.
1,-60 2,-30 3,-20 4,-15 5,-12 6,-10
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 -60.
1-60=-59 2-30=-28 3-20=-17 4-15=-11 5-12=-7 6-10=-4
Calculate the sum for each pair.
a=-12 b=5
The solution is the pair that gives sum -7.
\left(x^{2}-12x\right)+\left(5x-60\right)
Rewrite x^{2}-7x-60 as \left(x^{2}-12x\right)+\left(5x-60\right).
x\left(x-12\right)+5\left(x-12\right)
Factor out x in the first and 5 in the second group.
\left(x-12\right)\left(x+5\right)
Factor out common term x-12 by using distributive property.
x=12 x=-5
To find equation solutions, solve x-12=0 and x+5=0.
8x^{2}-56x-480=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(-56\right)±\sqrt{\left(-56\right)^{2}-4\times 8\left(-480\right)}}{2\times 8}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 8 for a, -56 for b, and -480 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-56\right)±\sqrt{3136-4\times 8\left(-480\right)}}{2\times 8}
Square -56.
x=\frac{-\left(-56\right)±\sqrt{3136-32\left(-480\right)}}{2\times 8}
Multiply -4 times 8.
x=\frac{-\left(-56\right)±\sqrt{3136+15360}}{2\times 8}
Multiply -32 times -480.
x=\frac{-\left(-56\right)±\sqrt{18496}}{2\times 8}
Add 3136 to 15360.
x=\frac{-\left(-56\right)±136}{2\times 8}
Take the square root of 18496.
x=\frac{56±136}{2\times 8}
The opposite of -56 is 56.
x=\frac{56±136}{16}
Multiply 2 times 8.
x=\frac{192}{16}
Now solve the equation x=\frac{56±136}{16} when ± is plus. Add 56 to 136.
x=12
Divide 192 by 16.
x=-\frac{80}{16}
Now solve the equation x=\frac{56±136}{16} when ± is minus. Subtract 136 from 56.
x=-5
Divide -80 by 16.
x=12 x=-5
The equation is now solved.
8x^{2}-56x-480=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.
8x^{2}-56x-480-\left(-480\right)=-\left(-480\right)
Add 480 to both sides of the equation.
8x^{2}-56x=-\left(-480\right)
Subtracting -480 from itself leaves 0.
8x^{2}-56x=480
Subtract -480 from 0.
\frac{8x^{2}-56x}{8}=\frac{480}{8}
Divide both sides by 8.
x^{2}+\left(-\frac{56}{8}\right)x=\frac{480}{8}
Dividing by 8 undoes the multiplication by 8.
x^{2}-7x=\frac{480}{8}
Divide -56 by 8.
x^{2}-7x=60
Divide 480 by 8.
x^{2}-7x+\left(-\frac{7}{2}\right)^{2}=60+\left(-\frac{7}{2}\right)^{2}
Divide -7, the coefficient of the x term, by 2 to get -\frac{7}{2}. Then add the square of -\frac{7}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-7x+\frac{49}{4}=60+\frac{49}{4}
Square -\frac{7}{2} by squaring both the numerator and the denominator of the fraction.
x^{2}-7x+\frac{49}{4}=\frac{289}{4}
Add 60 to \frac{49}{4}.
\left(x-\frac{7}{2}\right)^{2}=\frac{289}{4}
Factor x^{2}-7x+\frac{49}{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{7}{2}\right)^{2}}=\sqrt{\frac{289}{4}}
Take the square root of both sides of the equation.
x-\frac{7}{2}=\frac{17}{2} x-\frac{7}{2}=-\frac{17}{2}
Simplify.
x=12 x=-5
Add \frac{7}{2} to both sides of the equation.
x ^ 2 -7x -60 = 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.This is achieved by dividing both sides of the equation by 8
r + s = 7 rs = -60
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{7}{2} - u s = \frac{7}{2} + u
Two numbers r and s sum up to 7 exactly when the average of the two numbers is \frac{1}{2}*7 = \frac{7}{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{7}{2} - u) (\frac{7}{2} + u) = -60
To solve for unknown quantity u, substitute these in the product equation rs = -60
\frac{49}{4} - u^2 = -60
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -60-\frac{49}{4} = -\frac{289}{4}
Simplify the expression by subtracting \frac{49}{4} on both sides
u^2 = \frac{289}{4} u = \pm\sqrt{\frac{289}{4}} = \pm \frac{17}{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{7}{2} - \frac{17}{2} = -5 s = \frac{7}{2} + \frac{17}{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.
Examples
Quadratic equation
{ x } ^ { 2 } - 4 x - 5 = 0
Trigonometry
4 \sin \theta \cos \theta = 2 \sin \theta
Linear equation
y = 3x + 4
Arithmetic
699 * 533
Matrix
\left[ \begin{array} { l l } { 2 } & { 3 } \\ { 5 } & { 4 } \end{array} \right] \left[ \begin{array} { l l l } { 2 } & { 0 } & { 3 } \\ { -1 } & { 1 } & { 5 } \end{array} \right]
Simultaneous equation
\left. \begin{cases} { 8x+2y = 46 } \\ { 7x+3y = 47 } \end{cases} \right.
Differentiation
\frac { d } { d x } \frac { ( 3 x ^ { 2 } - 2 ) } { ( x - 5 ) }
Integration
\int _ { 0 } ^ { 1 } x e ^ { - x ^ { 2 } } d x
Limits
\lim _{x \rightarrow-3} \frac{x^{2}-9}{x^{2}+2 x-3}