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x^{2}+5x+9-5=0
Subtract 5 from both sides.
x^{2}+5x+4=0
Subtract 5 from 9 to get 4.
a+b=5 ab=4
To solve the equation, factor x^{2}+5x+4 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). 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 positive, a and b are both positive. List all such integer pairs that give product 4.
1+4=5 2+2=4
Calculate the sum for each pair.
a=1 b=4
The solution is the pair that gives sum 5.
\left(x+1\right)\left(x+4\right)
Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.
x=-1 x=-4
To find equation solutions, solve x+1=0 and x+4=0.
x^{2}+5x+9-5=0
Subtract 5 from both sides.
x^{2}+5x+4=0
Subtract 5 from 9 to get 4.
a+b=5 ab=1\times 4=4
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+4. 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 positive, a and b are both positive. List all such integer pairs that give product 4.
1+4=5 2+2=4
Calculate the sum for each pair.
a=1 b=4
The solution is the pair that gives sum 5.
\left(x^{2}+x\right)+\left(4x+4\right)
Rewrite x^{2}+5x+4 as \left(x^{2}+x\right)+\left(4x+4\right).
x\left(x+1\right)+4\left(x+1\right)
Factor out x in the first and 4 in the second group.
\left(x+1\right)\left(x+4\right)
Factor out common term x+1 by using distributive property.
x=-1 x=-4
To find equation solutions, solve x+1=0 and x+4=0.
x^{2}+5x+9=5
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^{2}+5x+9-5=5-5
Subtract 5 from both sides of the equation.
x^{2}+5x+9-5=0
Subtracting 5 from itself leaves 0.
x^{2}+5x+4=0
Subtract 5 from 9.
x=\frac{-5±\sqrt{5^{2}-4\times 4}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 5 for b, and 4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-5±\sqrt{25-4\times 4}}{2}
Square 5.
x=\frac{-5±\sqrt{25-16}}{2}
Multiply -4 times 4.
x=\frac{-5±\sqrt{9}}{2}
Add 25 to -16.
x=\frac{-5±3}{2}
Take the square root of 9.
x=-\frac{2}{2}
Now solve the equation x=\frac{-5±3}{2} when ± is plus. Add -5 to 3.
x=-1
Divide -2 by 2.
x=-\frac{8}{2}
Now solve the equation x=\frac{-5±3}{2} when ± is minus. Subtract 3 from -5.
x=-4
Divide -8 by 2.
x=-1 x=-4
The equation is now solved.
x^{2}+5x+9=5
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.
x^{2}+5x+9-9=5-9
Subtract 9 from both sides of the equation.
x^{2}+5x=5-9
Subtracting 9 from itself leaves 0.
x^{2}+5x=-4
Subtract 9 from 5.
x^{2}+5x+\left(\frac{5}{2}\right)^{2}=-4+\left(\frac{5}{2}\right)^{2}
Divide 5, the coefficient of the x term, by 2 to get \frac{5}{2}. Then add the square of \frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+5x+\frac{25}{4}=-4+\frac{25}{4}
Square \frac{5}{2} by squaring both the numerator and the denominator of the fraction.
x^{2}+5x+\frac{25}{4}=\frac{9}{4}
Add -4 to \frac{25}{4}.
\left(x+\frac{5}{2}\right)^{2}=\frac{9}{4}
Factor x^{2}+5x+\frac{25}{4}. 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}{2}\right)^{2}}=\sqrt{\frac{9}{4}}
Take the square root of both sides of the equation.
x+\frac{5}{2}=\frac{3}{2} x+\frac{5}{2}=-\frac{3}{2}
Simplify.
x=-1 x=-4
Subtract \frac{5}{2} from both sides of the equation.