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5x^{2}+1+13x=-5
Add 13x to both sides.
5x^{2}+1+13x+5=0
Add 5 to both sides.
5x^{2}+6+13x=0
Add 1 and 5 to get 6.
5x^{2}+13x+6=0
Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.
a+b=13 ab=5\times 6=30
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 5x^{2}+ax+bx+6. To find a and b, set up a system to be solved.
1,30 2,15 3,10 5,6
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 30.
1+30=31 2+15=17 3+10=13 5+6=11
Calculate the sum for each pair.
a=3 b=10
The solution is the pair that gives sum 13.
\left(5x^{2}+3x\right)+\left(10x+6\right)
Rewrite 5x^{2}+13x+6 as \left(5x^{2}+3x\right)+\left(10x+6\right).
x\left(5x+3\right)+2\left(5x+3\right)
Factor out x in the first and 2 in the second group.
\left(5x+3\right)\left(x+2\right)
Factor out common term 5x+3 by using distributive property.
x=-\frac{3}{5} x=-2
To find equation solutions, solve 5x+3=0 and x+2=0.
5x^{2}+1+13x=-5
Add 13x to both sides.
5x^{2}+1+13x+5=0
Add 5 to both sides.
5x^{2}+6+13x=0
Add 1 and 5 to get 6.
5x^{2}+13x+6=0
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=\frac{-13±\sqrt{13^{2}-4\times 5\times 6}}{2\times 5}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, 13 for b, and 6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-13±\sqrt{169-4\times 5\times 6}}{2\times 5}
Square 13.
x=\frac{-13±\sqrt{169-20\times 6}}{2\times 5}
Multiply -4 times 5.
x=\frac{-13±\sqrt{169-120}}{2\times 5}
Multiply -20 times 6.
x=\frac{-13±\sqrt{49}}{2\times 5}
Add 169 to -120.
x=\frac{-13±7}{2\times 5}
Take the square root of 49.
x=\frac{-13±7}{10}
Multiply 2 times 5.
x=-\frac{6}{10}
Now solve the equation x=\frac{-13±7}{10} when ± is plus. Add -13 to 7.
x=-\frac{3}{5}
Reduce the fraction \frac{-6}{10} to lowest terms by extracting and canceling out 2.
x=-\frac{20}{10}
Now solve the equation x=\frac{-13±7}{10} when ± is minus. Subtract 7 from -13.
x=-2
Divide -20 by 10.
x=-\frac{3}{5} x=-2
The equation is now solved.
5x^{2}+1+13x=-5
Add 13x to both sides.
5x^{2}+13x=-5-1
Subtract 1 from both sides.
5x^{2}+13x=-6
Subtract 1 from -5 to get -6.
\frac{5x^{2}+13x}{5}=-\frac{6}{5}
Divide both sides by 5.
x^{2}+\frac{13}{5}x=-\frac{6}{5}
Dividing by 5 undoes the multiplication by 5.
x^{2}+\frac{13}{5}x+\left(\frac{13}{10}\right)^{2}=-\frac{6}{5}+\left(\frac{13}{10}\right)^{2}
Divide \frac{13}{5}, the coefficient of the x term, by 2 to get \frac{13}{10}. Then add the square of \frac{13}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+\frac{13}{5}x+\frac{169}{100}=-\frac{6}{5}+\frac{169}{100}
Square \frac{13}{10} by squaring both the numerator and the denominator of the fraction.
x^{2}+\frac{13}{5}x+\frac{169}{100}=\frac{49}{100}
Add -\frac{6}{5} to \frac{169}{100} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x+\frac{13}{10}\right)^{2}=\frac{49}{100}
Factor x^{2}+\frac{13}{5}x+\frac{169}{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{13}{10}\right)^{2}}=\sqrt{\frac{49}{100}}
Take the square root of both sides of the equation.
x+\frac{13}{10}=\frac{7}{10} x+\frac{13}{10}=-\frac{7}{10}
Simplify.
x=-\frac{3}{5} x=-2
Subtract \frac{13}{10} from both sides of the equation.