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x^{2}-x-6=0
Divide both sides by 5.
a+b=-1 ab=1\left(-6\right)=-6
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-6. To find a and b, set up a system to be solved.
1,-6 2,-3
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 -6.
1-6=-5 2-3=-1
Calculate the sum for each pair.
a=-3 b=2
The solution is the pair that gives sum -1.
\left(x^{2}-3x\right)+\left(2x-6\right)
Rewrite x^{2}-x-6 as \left(x^{2}-3x\right)+\left(2x-6\right).
x\left(x-3\right)+2\left(x-3\right)
Factor out x in the first and 2 in the second group.
\left(x-3\right)\left(x+2\right)
Factor out common term x-3 by using distributive property.
x=3 x=-2
To find equation solutions, solve x-3=0 and x+2=0.
5x^{2}-5x-30=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(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 5\left(-30\right)}}{2\times 5}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -5 for b, and -30 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-5\right)±\sqrt{25-4\times 5\left(-30\right)}}{2\times 5}
Square -5.
x=\frac{-\left(-5\right)±\sqrt{25-20\left(-30\right)}}{2\times 5}
Multiply -4 times 5.
x=\frac{-\left(-5\right)±\sqrt{25+600}}{2\times 5}
Multiply -20 times -30.
x=\frac{-\left(-5\right)±\sqrt{625}}{2\times 5}
Add 25 to 600.
x=\frac{-\left(-5\right)±25}{2\times 5}
Take the square root of 625.
x=\frac{5±25}{2\times 5}
The opposite of -5 is 5.
x=\frac{5±25}{10}
Multiply 2 times 5.
x=\frac{30}{10}
Now solve the equation x=\frac{5±25}{10} when ± is plus. Add 5 to 25.
x=3
Divide 30 by 10.
x=-\frac{20}{10}
Now solve the equation x=\frac{5±25}{10} when ± is minus. Subtract 25 from 5.
x=-2
Divide -20 by 10.
x=3 x=-2
The equation is now solved.
5x^{2}-5x-30=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.
5x^{2}-5x-30-\left(-30\right)=-\left(-30\right)
Add 30 to both sides of the equation.
5x^{2}-5x=-\left(-30\right)
Subtracting -30 from itself leaves 0.
5x^{2}-5x=30
Subtract -30 from 0.
\frac{5x^{2}-5x}{5}=\frac{30}{5}
Divide both sides by 5.
x^{2}+\left(-\frac{5}{5}\right)x=\frac{30}{5}
Dividing by 5 undoes the multiplication by 5.
x^{2}-x=\frac{30}{5}
Divide -5 by 5.
x^{2}-x=6
Divide 30 by 5.
x^{2}-x+\left(-\frac{1}{2}\right)^{2}=6+\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}=6+\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{25}{4}
Add 6 to \frac{1}{4}.
\left(x-\frac{1}{2}\right)^{2}=\frac{25}{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{25}{4}}
Take the square root of both sides of the equation.
x-\frac{1}{2}=\frac{5}{2} x-\frac{1}{2}=-\frac{5}{2}
Simplify.
x=3 x=-2
Add \frac{1}{2} to both sides of the equation.
x ^ 2 -1x -6 = 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 5
r + s = 1 rs = -6
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) = -6
To solve for unknown quantity u, substitute these in the product equation rs = -6
\frac{1}{4} - u^2 = -6
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -6-\frac{1}{4} = -\frac{25}{4}
Simplify the expression by subtracting \frac{1}{4} on both sides
u^2 = \frac{25}{4} u = \pm\sqrt{\frac{25}{4}} = \pm \frac{5}{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{5}{2} = -2 s = \frac{1}{2} + \frac{5}{2} = 3
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.