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n^{2}+7n+15-5=0
Subtract 5 from both sides.
n^{2}+7n+10=0
Subtract 5 from 15 to get 10.
a+b=7 ab=10
To solve the equation, factor n^{2}+7n+10 using formula n^{2}+\left(a+b\right)n+ab=\left(n+a\right)\left(n+b\right). To find a and b, set up a system to be solved.
1,10 2,5
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 10.
1+10=11 2+5=7
Calculate the sum for each pair.
a=2 b=5
The solution is the pair that gives sum 7.
\left(n+2\right)\left(n+5\right)
Rewrite factored expression \left(n+a\right)\left(n+b\right) using the obtained values.
n=-2 n=-5
To find equation solutions, solve n+2=0 and n+5=0.
n^{2}+7n+15-5=0
Subtract 5 from both sides.
n^{2}+7n+10=0
Subtract 5 from 15 to get 10.
a+b=7 ab=1\times 10=10
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as n^{2}+an+bn+10. To find a and b, set up a system to be solved.
1,10 2,5
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 10.
1+10=11 2+5=7
Calculate the sum for each pair.
a=2 b=5
The solution is the pair that gives sum 7.
\left(n^{2}+2n\right)+\left(5n+10\right)
Rewrite n^{2}+7n+10 as \left(n^{2}+2n\right)+\left(5n+10\right).
n\left(n+2\right)+5\left(n+2\right)
Factor out n in the first and 5 in the second group.
\left(n+2\right)\left(n+5\right)
Factor out common term n+2 by using distributive property.
n=-2 n=-5
To find equation solutions, solve n+2=0 and n+5=0.
n^{2}+7n+15=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.
n^{2}+7n+15-5=5-5
Subtract 5 from both sides of the equation.
n^{2}+7n+15-5=0
Subtracting 5 from itself leaves 0.
n^{2}+7n+10=0
Subtract 5 from 15.
n=\frac{-7±\sqrt{7^{2}-4\times 10}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 7 for b, and 10 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
n=\frac{-7±\sqrt{49-4\times 10}}{2}
Square 7.
n=\frac{-7±\sqrt{49-40}}{2}
Multiply -4 times 10.
n=\frac{-7±\sqrt{9}}{2}
Add 49 to -40.
n=\frac{-7±3}{2}
Take the square root of 9.
n=-\frac{4}{2}
Now solve the equation n=\frac{-7±3}{2} when ± is plus. Add -7 to 3.
n=-2
Divide -4 by 2.
n=-\frac{10}{2}
Now solve the equation n=\frac{-7±3}{2} when ± is minus. Subtract 3 from -7.
n=-5
Divide -10 by 2.
n=-2 n=-5
The equation is now solved.
n^{2}+7n+15=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.
n^{2}+7n+15-15=5-15
Subtract 15 from both sides of the equation.
n^{2}+7n=5-15
Subtracting 15 from itself leaves 0.
n^{2}+7n=-10
Subtract 15 from 5.
n^{2}+7n+\left(\frac{7}{2}\right)^{2}=-10+\left(\frac{7}{2}\right)^{2}
Divide 7, the coefficient of the x term, by 2 to get \frac{7}{2}. Then add the square of \frac{7}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
n^{2}+7n+\frac{49}{4}=-10+\frac{49}{4}
Square \frac{7}{2} by squaring both the numerator and the denominator of the fraction.
n^{2}+7n+\frac{49}{4}=\frac{9}{4}
Add -10 to \frac{49}{4}.
\left(n+\frac{7}{2}\right)^{2}=\frac{9}{4}
Factor n^{2}+7n+\frac{49}{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(n+\frac{7}{2}\right)^{2}}=\sqrt{\frac{9}{4}}
Take the square root of both sides of the equation.
n+\frac{7}{2}=\frac{3}{2} n+\frac{7}{2}=-\frac{3}{2}
Simplify.
n=-2 n=-5
Subtract \frac{7}{2} from both sides of the equation.