Solve for x

Steps Using Factoring
Steps Using Factoring By Grouping
Steps for Completing the Square
Graph
Graph Both Sides in 2D
Graph in 2D
Still have questions?
x^{2}+x-20=0
Subtract 20 from both sides.
a+b=1 ab=-20
To solve the equation, factor x^{2}+x-20 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.
-1,20 -2,10 -4,5
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 -20.
-1+20=19 -2+10=8 -4+5=1
Calculate the sum for each pair.
a=-4 b=5
The solution is the pair that gives sum 1.
\left(x-4\right)\left(x+5\right)
Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.
x=4 x=-5
To find equation solutions, solve x-4=0 and x+5=0.
x^{2}+x-20=0
Subtract 20 from both sides.
a+b=1 ab=1\left(-20\right)=-20
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-20. To find a and b, set up a system to be solved.
-1,20 -2,10 -4,5
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 -20.
-1+20=19 -2+10=8 -4+5=1
Calculate the sum for each pair.
a=-4 b=5
The solution is the pair that gives sum 1.
\left(x^{2}-4x\right)+\left(5x-20\right)
Rewrite x^{2}+x-20 as \left(x^{2}-4x\right)+\left(5x-20\right).
x\left(x-4\right)+5\left(x-4\right)
Factor out x in the first and 5 in the second group.
\left(x-4\right)\left(x+5\right)
Factor out common term x-4 by using distributive property.
x=4 x=-5
To find equation solutions, solve x-4=0 and x+5=0.
x^{2}+x=20
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^{2}+x-20=20-20
Subtract 20 from both sides of the equation.
x^{2}+x-20=0
Subtracting 20 from itself leaves 0.
x=\frac{-1±\sqrt{1^{2}-4\left(-20\right)}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 1 for b, and -20 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-1±\sqrt{1-4\left(-20\right)}}{2}
Square 1.
x=\frac{-1±\sqrt{1+80}}{2}
Multiply -4 times -20.
x=\frac{-1±\sqrt{81}}{2}
x=\frac{-1±9}{2}
Take the square root of 81.
x=\frac{8}{2}
Now solve the equation x=\frac{-1±9}{2} when ± is plus. Add -1 to 9.
x=4
Divide 8 by 2.
x=\frac{-10}{2}
Now solve the equation x=\frac{-1±9}{2} when ± is minus. Subtract 9 from -1.
x=-5
Divide -10 by 2.
x=4 x=-5
The equation is now solved.
x^{2}+x=20
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}+x+\left(\frac{1}{2}\right)^{2}=20+\left(\frac{1}{2}\right)^{2}
Divide 1, the coefficient of the x term, by 2 to get \frac{1}{2}=0.5. Then add the square of \frac{1}{2}=0.5 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}=20+\frac{1}{4}
Square \frac{1}{2}=0.5 by squaring both the numerator and the denominator of the fraction.
x^{2}+x+\frac{1}{4}=\frac{81}{4}