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5a^{2}-26a-24=0
Subtract 24 from both sides.
a+b=-26 ab=5\left(-24\right)=-120
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 5a^{2}+aa+ba-24. To find a and b, set up a system to be solved.
1,-120 2,-60 3,-40 4,-30 5,-24 6,-20 8,-15 10,-12
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 -120.
1-120=-119 2-60=-58 3-40=-37 4-30=-26 5-24=-19 6-20=-14 8-15=-7 10-12=-2
Calculate the sum for each pair.
a=-30 b=4
The solution is the pair that gives sum -26.
\left(5a^{2}-30a\right)+\left(4a-24\right)
Rewrite 5a^{2}-26a-24 as \left(5a^{2}-30a\right)+\left(4a-24\right).
5a\left(a-6\right)+4\left(a-6\right)
Factor out 5a in the first and 4 in the second group.
\left(a-6\right)\left(5a+4\right)
Factor out common term a-6 by using distributive property.
a=6 a=-\frac{4}{5}
To find equation solutions, solve a-6=0 and 5a+4=0.
5a^{2}-26a=24
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.
5a^{2}-26a-24=24-24
Subtract 24 from both sides of the equation.
5a^{2}-26a-24=0
Subtracting 24 from itself leaves 0.
a=\frac{-\left(-26\right)±\sqrt{\left(-26\right)^{2}-4\times 5\left(-24\right)}}{2\times 5}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -26 for b, and -24 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
a=\frac{-\left(-26\right)±\sqrt{676-4\times 5\left(-24\right)}}{2\times 5}
Square -26.
a=\frac{-\left(-26\right)±\sqrt{676-20\left(-24\right)}}{2\times 5}
Multiply -4 times 5.
a=\frac{-\left(-26\right)±\sqrt{676+480}}{2\times 5}
Multiply -20 times -24.
a=\frac{-\left(-26\right)±\sqrt{1156}}{2\times 5}
Add 676 to 480.
a=\frac{-\left(-26\right)±34}{2\times 5}
Take the square root of 1156.
a=\frac{26±34}{2\times 5}
The opposite of -26 is 26.
a=\frac{26±34}{10}
Multiply 2 times 5.
a=\frac{60}{10}
Now solve the equation a=\frac{26±34}{10} when ± is plus. Add 26 to 34.
a=6
Divide 60 by 10.
a=-\frac{8}{10}
Now solve the equation a=\frac{26±34}{10} when ± is minus. Subtract 34 from 26.
a=-\frac{4}{5}
Reduce the fraction \frac{-8}{10} to lowest terms by extracting and canceling out 2.
a=6 a=-\frac{4}{5}
The equation is now solved.
5a^{2}-26a=24
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{5a^{2}-26a}{5}=\frac{24}{5}
Divide both sides by 5.
a^{2}-\frac{26}{5}a=\frac{24}{5}
Dividing by 5 undoes the multiplication by 5.
a^{2}-\frac{26}{5}a+\left(-\frac{13}{5}\right)^{2}=\frac{24}{5}+\left(-\frac{13}{5}\right)^{2}
Divide -\frac{26}{5}, the coefficient of the x term, by 2 to get -\frac{13}{5}. Then add the square of -\frac{13}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
a^{2}-\frac{26}{5}a+\frac{169}{25}=\frac{24}{5}+\frac{169}{25}
Square -\frac{13}{5} by squaring both the numerator and the denominator of the fraction.
a^{2}-\frac{26}{5}a+\frac{169}{25}=\frac{289}{25}
Add \frac{24}{5} to \frac{169}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(a-\frac{13}{5}\right)^{2}=\frac{289}{25}
Factor a^{2}-\frac{26}{5}a+\frac{169}{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(a-\frac{13}{5}\right)^{2}}=\sqrt{\frac{289}{25}}
Take the square root of both sides of the equation.
a-\frac{13}{5}=\frac{17}{5} a-\frac{13}{5}=-\frac{17}{5}
Simplify.
a=6 a=-\frac{4}{5}
Add \frac{13}{5} to both sides of the equation.