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8x^{2}-20x+8=0
Combine -16x and -4x to get -20x.
2x^{2}-5x+2=0
Divide both sides by 4.
a+b=-5 ab=2\times 2=4
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2x^{2}+ax+bx+2. To find a and b, set up a system to be solved.
-1,-4 -2,-2
Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 4.
-1-4=-5 -2-2=-4
Calculate the sum for each pair.
a=-4 b=-1
The solution is the pair that gives sum -5.
\left(2x^{2}-4x\right)+\left(-x+2\right)
Rewrite 2x^{2}-5x+2 as \left(2x^{2}-4x\right)+\left(-x+2\right).
2x\left(x-2\right)-\left(x-2\right)
Factor out 2x in the first and -1 in the second group.
\left(x-2\right)\left(2x-1\right)
Factor out common term x-2 by using distributive property.
x=2 x=\frac{1}{2}
To find equation solutions, solve x-2=0 and 2x-1=0.
8x^{2}-20x+8=0
Combine -16x and -4x to get -20x.
x=\frac{-\left(-20\right)±\sqrt{\left(-20\right)^{2}-4\times 8\times 8}}{2\times 8}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 8 for a, -20 for b, and 8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-20\right)±\sqrt{400-4\times 8\times 8}}{2\times 8}
Square -20.
x=\frac{-\left(-20\right)±\sqrt{400-32\times 8}}{2\times 8}
Multiply -4 times 8.
x=\frac{-\left(-20\right)±\sqrt{400-256}}{2\times 8}
Multiply -32 times 8.
x=\frac{-\left(-20\right)±\sqrt{144}}{2\times 8}
Add 400 to -256.
x=\frac{-\left(-20\right)±12}{2\times 8}
Take the square root of 144.
x=\frac{20±12}{2\times 8}
The opposite of -20 is 20.
x=\frac{20±12}{16}
Multiply 2 times 8.
x=\frac{32}{16}
Now solve the equation x=\frac{20±12}{16} when ± is plus. Add 20 to 12.
x=2
Divide 32 by 16.
x=\frac{8}{16}
Now solve the equation x=\frac{20±12}{16} when ± is minus. Subtract 12 from 20.
x=\frac{1}{2}
Reduce the fraction \frac{8}{16} to lowest terms by extracting and canceling out 8.
x=2 x=\frac{1}{2}
The equation is now solved.
8x^{2}-20x+8=0
Combine -16x and -4x to get -20x.
8x^{2}-20x=-8
Subtract 8 from both sides. Anything subtracted from zero gives its negation.
\frac{8x^{2}-20x}{8}=-\frac{8}{8}
Divide both sides by 8.
x^{2}+\left(-\frac{20}{8}\right)x=-\frac{8}{8}
Dividing by 8 undoes the multiplication by 8.
x^{2}-\frac{5}{2}x=-\frac{8}{8}
Reduce the fraction \frac{-20}{8} to lowest terms by extracting and canceling out 4.
x^{2}-\frac{5}{2}x=-1
Divide -8 by 8.
x^{2}-\frac{5}{2}x+\left(-\frac{5}{4}\right)^{2}=-1+\left(-\frac{5}{4}\right)^{2}
Divide -\frac{5}{2}, the coefficient of the x term, by 2 to get -\frac{5}{4}. Then add the square of -\frac{5}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{5}{2}x+\frac{25}{16}=-1+\frac{25}{16}
Square -\frac{5}{4} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{5}{2}x+\frac{25}{16}=\frac{9}{16}
Add -1 to \frac{25}{16}.
\left(x-\frac{5}{4}\right)^{2}=\frac{9}{16}
Factor x^{2}-\frac{5}{2}x+\frac{25}{16}. 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{5}{4}\right)^{2}}=\sqrt{\frac{9}{16}}
Take the square root of both sides of the equation.
x-\frac{5}{4}=\frac{3}{4} x-\frac{5}{4}=-\frac{3}{4}
Simplify.
x=2 x=\frac{1}{2}
Add \frac{5}{4} to both sides of the equation.