Solve 5n^2=25-20n

$$5 n ^ { 2 } = 25 - 20 n$$

Solve for n

$$n=-5$$

$$n=1$$

Steps Using Factoring By Grouping

Subtract 25 from both sides.

$$5n^{2}-25=-20n$$

Add 20n to both sides.

$$5n^{2}-25+20n=0$$

Divide both sides by 5.

$$n^{2}-5+4n=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$n^{2}+4n-5=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as n^{2}+an+bn-5. To find a and b, set up a system to be solved.

$$a+b=4 ab=1\left(-5\right)=-5$$

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. The only such pair is the system solution.

$$a=-1 b=5$$

Rewrite n^{2}+4n-5 as \left(n^{2}-n\right)+\left(5n-5\right).

$$\left(n^{2}-n\right)+\left(5n-5\right)$$

Factor out n in the first and 5 in the second group.

$$n\left(n-1\right)+5\left(n-1\right)$$

Factor out common term n-1 by using distributive property.

$$\left(n-1\right)\left(n+5\right)$$

To find equation solutions, solve n-1=0 and n+5=0.

$$n=1 n=-5$$

Steps Using the Quadratic Formula

Subtract 25 from both sides.

$$5n^{2}-25=-20n$$

Add 20n to both sides.

$$5n^{2}-25+20n=0$$

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.

$$5n^{2}+20n-25=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, 20 for b, and -25 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-20±\sqrt{20^{2}-4\times 5\left(-25\right)}}{2\times 5}$$

Square 20.

$$n=\frac{-20±\sqrt{400-4\times 5\left(-25\right)}}{2\times 5}$$

Multiply -4 times 5.

$$n=\frac{-20±\sqrt{400-20\left(-25\right)}}{2\times 5}$$

Multiply -20 times -25.

$$n=\frac{-20±\sqrt{400+500}}{2\times 5}$$

Add 400 to 500.

$$n=\frac{-20±\sqrt{900}}{2\times 5}$$

Take the square root of 900.

$$n=\frac{-20±30}{2\times 5}$$

Multiply 2 times 5.

$$n=\frac{-20±30}{10}$$

Now solve the equation n=\frac{-20±30}{10} when ± is plus. Add -20 to 30.

$$n=\frac{10}{10}$$

Divide 10 by 10.

$$n=1$$

Now solve the equation n=\frac{-20±30}{10} when ± is minus. Subtract 30 from -20.

$$n=-\frac{50}{10}$$

Divide -50 by 10.

$$n=-5$$

The equation is now solved.

$$n=1 n=-5$$

Steps for Completing the Square

Add 20n to both sides.

$$5n^{2}+20n=25$$

Divide both sides by 5.

$$\frac{5n^{2}+20n}{5}=\frac{25}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$n^{2}+\frac{20}{5}n=\frac{25}{5}$$

Divide 20 by 5.

$$n^{2}+4n=\frac{25}{5}$$

Divide 25 by 5.

$$n^{2}+4n=5$$

Divide 4, the coefficient of the x term, by 2 to get 2. Then add the square of 2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$n^{2}+4n+2^{2}=5+2^{2}$$

Square 2.

$$n^{2}+4n+4=5+4$$

Add 5 to 4.

$$n^{2}+4n+4=9$$

Factor n^{2}+4n+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}.

$$\left(n+2\right)^{2}=9$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n+2\right)^{2}}=\sqrt{9}$$

Simplify.

$$n+2=3 n+2=-3$$

Subtract 2 from both sides of the equation.

$$n=1 n=-5$$