Skip to main content
Solve for n
Tick mark Image

Similar Problems from Web Search

Share

n^{2}+5n-500=0
Subtract 500 from both sides.
a+b=5 ab=-500
To solve the equation, factor n^{2}+5n-500 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.
-1,500 -2,250 -4,125 -5,100 -10,50 -20,25
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 -500.
-1+500=499 -2+250=248 -4+125=121 -5+100=95 -10+50=40 -20+25=5
Calculate the sum for each pair.
a=-20 b=25
The solution is the pair that gives sum 5.
\left(n-20\right)\left(n+25\right)
Rewrite factored expression \left(n+a\right)\left(n+b\right) using the obtained values.
n=20 n=-25
To find equation solutions, solve n-20=0 and n+25=0.
n^{2}+5n-500=0
Subtract 500 from both sides.
a+b=5 ab=1\left(-500\right)=-500
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as n^{2}+an+bn-500. To find a and b, set up a system to be solved.
-1,500 -2,250 -4,125 -5,100 -10,50 -20,25
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 -500.
-1+500=499 -2+250=248 -4+125=121 -5+100=95 -10+50=40 -20+25=5
Calculate the sum for each pair.
a=-20 b=25
The solution is the pair that gives sum 5.
\left(n^{2}-20n\right)+\left(25n-500\right)
Rewrite n^{2}+5n-500 as \left(n^{2}-20n\right)+\left(25n-500\right).
n\left(n-20\right)+25\left(n-20\right)
Factor out n in the first and 25 in the second group.
\left(n-20\right)\left(n+25\right)
Factor out common term n-20 by using distributive property.
n=20 n=-25
To find equation solutions, solve n-20=0 and n+25=0.
n^{2}+5n=500
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}+5n-500=500-500
Subtract 500 from both sides of the equation.
n^{2}+5n-500=0
Subtracting 500 from itself leaves 0.
n=\frac{-5±\sqrt{5^{2}-4\left(-500\right)}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 5 for b, and -500 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
n=\frac{-5±\sqrt{25-4\left(-500\right)}}{2}
Square 5.
n=\frac{-5±\sqrt{25+2000}}{2}
Multiply -4 times -500.
n=\frac{-5±\sqrt{2025}}{2}
Add 25 to 2000.
n=\frac{-5±45}{2}
Take the square root of 2025.
n=\frac{40}{2}
Now solve the equation n=\frac{-5±45}{2} when ± is plus. Add -5 to 45.
n=20
Divide 40 by 2.
n=-\frac{50}{2}
Now solve the equation n=\frac{-5±45}{2} when ± is minus. Subtract 45 from -5.
n=-25
Divide -50 by 2.
n=20 n=-25
The equation is now solved.
n^{2}+5n=500
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}+5n+\left(\frac{5}{2}\right)^{2}=500+\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.
n^{2}+5n+\frac{25}{4}=500+\frac{25}{4}
Square \frac{5}{2} by squaring both the numerator and the denominator of the fraction.
n^{2}+5n+\frac{25}{4}=\frac{2025}{4}
Add 500 to \frac{25}{4}.
\left(n+\frac{5}{2}\right)^{2}=\frac{2025}{4}
Factor n^{2}+5n+\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(n+\frac{5}{2}\right)^{2}}=\sqrt{\frac{2025}{4}}
Take the square root of both sides of the equation.
n+\frac{5}{2}=\frac{45}{2} n+\frac{5}{2}=-\frac{45}{2}
Simplify.
n=20 n=-25
Subtract \frac{5}{2} from both sides of the equation.