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-2x^{2}+20x-48=0
Subtract 48 from both sides.
-x^{2}+10x-24=0
Divide both sides by 2.
a+b=10 ab=-\left(-24\right)=24
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-24. To find a and b, set up a system to be solved.
1,24 2,12 3,8 4,6
Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 24.
1+24=25 2+12=14 3+8=11 4+6=10
Calculate the sum for each pair.
a=6 b=4
The solution is the pair that gives sum 10.
\left(-x^{2}+6x\right)+\left(4x-24\right)
Rewrite -x^{2}+10x-24 as \left(-x^{2}+6x\right)+\left(4x-24\right).
-x\left(x-6\right)+4\left(x-6\right)
Factor out -x in the first and 4 in the second group.
\left(x-6\right)\left(-x+4\right)
Factor out common term x-6 by using distributive property.
x=6 x=4
To find equation solutions, solve x-6=0 and -x+4=0.
-2x^{2}+20x=48
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.
-2x^{2}+20x-48=48-48
Subtract 48 from both sides of the equation.
-2x^{2}+20x-48=0
Subtracting 48 from itself leaves 0.
x=\frac{-20±\sqrt{20^{2}-4\left(-2\right)\left(-48\right)}}{2\left(-2\right)}
This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, 20 for b, and -48 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-20±\sqrt{400-4\left(-2\right)\left(-48\right)}}{2\left(-2\right)}
Square 20.
x=\frac{-20±\sqrt{400+8\left(-48\right)}}{2\left(-2\right)}
Multiply -4 times -2.
x=\frac{-20±\sqrt{400-384}}{2\left(-2\right)}
Multiply 8 times -48.
x=\frac{-20±\sqrt{16}}{2\left(-2\right)}
Add 400 to -384.
x=\frac{-20±4}{2\left(-2\right)}
Take the square root of 16.
x=\frac{-20±4}{-4}
Multiply 2 times -2.
x=-\frac{16}{-4}
Now solve the equation x=\frac{-20±4}{-4} when ± is plus. Add -20 to 4.
x=4
Divide -16 by -4.
x=-\frac{24}{-4}
Now solve the equation x=\frac{-20±4}{-4} when ± is minus. Subtract 4 from -20.
x=6
Divide -24 by -4.
x=4 x=6
The equation is now solved.
-2x^{2}+20x=48
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.
\frac{-2x^{2}+20x}{-2}=\frac{48}{-2}
Divide both sides by -2.
x^{2}+\frac{20}{-2}x=\frac{48}{-2}
Dividing by -2 undoes the multiplication by -2.
x^{2}-10x=\frac{48}{-2}
Divide 20 by -2.
x^{2}-10x=-24
Divide 48 by -2.
x^{2}-10x+\left(-5\right)^{2}=-24+\left(-5\right)^{2}
Divide -10, the coefficient of the x term, by 2 to get -5. Then add the square of -5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-10x+25=-24+25
Square -5.
x^{2}-10x+25=1
Add -24 to 25.
\left(x-5\right)^{2}=1
Factor x^{2}-10x+25. 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-5\right)^{2}}=\sqrt{1}
Take the square root of both sides of the equation.
x-5=1 x-5=-1
Simplify.
x=6 x=4
Add 5 to both sides of the equation.