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5x^{2}+23x-10=0
Subtract 10 from both sides.
a+b=23 ab=5\left(-10\right)=-50
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 5x^{2}+ax+bx-10. To find a and b, set up a system to be solved.
-1,50 -2,25 -5,10
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 -50.
-1+50=49 -2+25=23 -5+10=5
Calculate the sum for each pair.
a=-2 b=25
The solution is the pair that gives sum 23.
\left(5x^{2}-2x\right)+\left(25x-10\right)
Rewrite 5x^{2}+23x-10 as \left(5x^{2}-2x\right)+\left(25x-10\right).
x\left(5x-2\right)+5\left(5x-2\right)
Factor out x in the first and 5 in the second group.
\left(5x-2\right)\left(x+5\right)
Factor out common term 5x-2 by using distributive property.
x=\frac{2}{5} x=-5
To find equation solutions, solve 5x-2=0 and x+5=0.
5x^{2}+23x=10
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.
5x^{2}+23x-10=10-10
Subtract 10 from both sides of the equation.
5x^{2}+23x-10=0
Subtracting 10 from itself leaves 0.
x=\frac{-23±\sqrt{23^{2}-4\times 5\left(-10\right)}}{2\times 5}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, 23 for b, and -10 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-23±\sqrt{529-4\times 5\left(-10\right)}}{2\times 5}
Square 23.
x=\frac{-23±\sqrt{529-20\left(-10\right)}}{2\times 5}
Multiply -4 times 5.
x=\frac{-23±\sqrt{529+200}}{2\times 5}
Multiply -20 times -10.
x=\frac{-23±\sqrt{729}}{2\times 5}
Add 529 to 200.
x=\frac{-23±27}{2\times 5}
Take the square root of 729.
x=\frac{-23±27}{10}
Multiply 2 times 5.
x=\frac{4}{10}
Now solve the equation x=\frac{-23±27}{10} when ± is plus. Add -23 to 27.
x=\frac{2}{5}
Reduce the fraction \frac{4}{10} to lowest terms by extracting and canceling out 2.
x=-\frac{50}{10}
Now solve the equation x=\frac{-23±27}{10} when ± is minus. Subtract 27 from -23.
x=-5
Divide -50 by 10.
x=\frac{2}{5} x=-5
The equation is now solved.
5x^{2}+23x=10
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{5x^{2}+23x}{5}=\frac{10}{5}
Divide both sides by 5.
x^{2}+\frac{23}{5}x=\frac{10}{5}
Dividing by 5 undoes the multiplication by 5.
x^{2}+\frac{23}{5}x=2
Divide 10 by 5.
x^{2}+\frac{23}{5}x+\left(\frac{23}{10}\right)^{2}=2+\left(\frac{23}{10}\right)^{2}
Divide \frac{23}{5}, the coefficient of the x term, by 2 to get \frac{23}{10}. Then add the square of \frac{23}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+\frac{23}{5}x+\frac{529}{100}=2+\frac{529}{100}
Square \frac{23}{10} by squaring both the numerator and the denominator of the fraction.
x^{2}+\frac{23}{5}x+\frac{529}{100}=\frac{729}{100}
Add 2 to \frac{529}{100}.
\left(x+\frac{23}{10}\right)^{2}=\frac{729}{100}
Factor x^{2}+\frac{23}{5}x+\frac{529}{100}. 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{23}{10}\right)^{2}}=\sqrt{\frac{729}{100}}
Take the square root of both sides of the equation.
x+\frac{23}{10}=\frac{27}{10} x+\frac{23}{10}=-\frac{27}{10}
Simplify.
x=\frac{2}{5} x=-5
Subtract \frac{23}{10} from both sides of the equation.