Solve for x
x=-3
x=4
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x^{2}-x-12=0
Divide both sides by 2.
a+b=-1 ab=1\left(-12\right)=-12
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-12. To find a and b, set up a system to be solved.
1,-12 2,-6 3,-4
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 -12.
1-12=-11 2-6=-4 3-4=-1
Calculate the sum for each pair.
a=-4 b=3
The solution is the pair that gives sum -1.
\left(x^{2}-4x\right)+\left(3x-12\right)
Rewrite x^{2}-x-12 as \left(x^{2}-4x\right)+\left(3x-12\right).
x\left(x-4\right)+3\left(x-4\right)
Factor out x in the first and 3 in the second group.
\left(x-4\right)\left(x+3\right)
Factor out common term x-4 by using distributive property.
x=4 x=-3
To find equation solutions, solve x-4=0 and x+3=0.
2x^{2}-2x-24=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(-2\right)±\sqrt{\left(-2\right)^{2}-4\times 2\left(-24\right)}}{2\times 2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -2 for b, and -24 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-2\right)±\sqrt{4-4\times 2\left(-24\right)}}{2\times 2}
Square -2.
x=\frac{-\left(-2\right)±\sqrt{4-8\left(-24\right)}}{2\times 2}
Multiply -4 times 2.
x=\frac{-\left(-2\right)±\sqrt{4+192}}{2\times 2}
Multiply -8 times -24.
x=\frac{-\left(-2\right)±\sqrt{196}}{2\times 2}
Add 4 to 192.
x=\frac{-\left(-2\right)±14}{2\times 2}
Take the square root of 196.
x=\frac{2±14}{2\times 2}
The opposite of -2 is 2.
x=\frac{2±14}{4}
Multiply 2 times 2.
x=\frac{16}{4}
Now solve the equation x=\frac{2±14}{4} when ± is plus. Add 2 to 14.
x=4
Divide 16 by 4.
x=-\frac{12}{4}
Now solve the equation x=\frac{2±14}{4} when ± is minus. Subtract 14 from 2.
x=-3
Divide -12 by 4.
x=4 x=-3
The equation is now solved.
2x^{2}-2x-24=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.
2x^{2}-2x-24-\left(-24\right)=-\left(-24\right)
Add 24 to both sides of the equation.
2x^{2}-2x=-\left(-24\right)
Subtracting -24 from itself leaves 0.
2x^{2}-2x=24
Subtract -24 from 0.
\frac{2x^{2}-2x}{2}=\frac{24}{2}
Divide both sides by 2.
x^{2}+\left(-\frac{2}{2}\right)x=\frac{24}{2}
Dividing by 2 undoes the multiplication by 2.
x^{2}-x=\frac{24}{2}
Divide -2 by 2.
x^{2}-x=12
Divide 24 by 2.
x^{2}-x+\left(-\frac{1}{2}\right)^{2}=12+\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}=12+\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{49}{4}
Add 12 to \frac{1}{4}.
\left(x-\frac{1}{2}\right)^{2}=\frac{49}{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{49}{4}}
Take the square root of both sides of the equation.
x-\frac{1}{2}=\frac{7}{2} x-\frac{1}{2}=-\frac{7}{2}
Simplify.
x=4 x=-3
Add \frac{1}{2} to both sides of the equation.
x ^ 2 -1x -12 = 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 2
r + s = 1 rs = -12
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{1}{2} - u s = \frac{1}{2} + u
Two numbers r and s sum up to 1 exactly when the average of the two numbers is \frac{1}{2}*1 = \frac{1}{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{1}{2} - u) (\frac{1}{2} + u) = -12
To solve for unknown quantity u, substitute these in the product equation rs = -12
\frac{1}{4} - u^2 = -12
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -12-\frac{1}{4} = -\frac{49}{4}
Simplify the expression by subtracting \frac{1}{4} on both sides
u^2 = \frac{49}{4} u = \pm\sqrt{\frac{49}{4}} = \pm \frac{7}{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{1}{2} - \frac{7}{2} = -3 s = \frac{1}{2} + \frac{7}{2} = 4
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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