Solve 4u^2+25=-20u

$$4 u ^ { 2 } + 25 = - 20 u$$

Solve for u

$$u = -\frac{5}{2} = -2\frac{1}{2} = -2.5$$

Steps Using Factoring By Grouping

Add 20u to both sides.

$$4u^{2}+25+20u=0$$

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

$$4u^{2}+20u+25=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 4u^{2}+au+bu+25. To find a and b, set up a system to be solved.

$$a+b=20 ab=4\times 25=100$$

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 100.

$$1,100 2,50 4,25 5,20 10,10$$

Calculate the sum for each pair.

$$1+100=101 2+50=52 4+25=29 5+20=25 10+10=20$$

The solution is the pair that gives sum 20.

$$a=10 b=10$$

Rewrite 4u^{2}+20u+25 as \left(4u^{2}+10u\right)+\left(10u+25\right).

$$\left(4u^{2}+10u\right)+\left(10u+25\right)$$

Factor out 2u in the first and 5 in the second group.

$$2u\left(2u+5\right)+5\left(2u+5\right)$$

Factor out common term 2u+5 by using distributive property.

$$\left(2u+5\right)\left(2u+5\right)$$

Rewrite as a binomial square.

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

To find equation solution, solve 2u+5=0.

$$u=-\frac{5}{2}$$

Steps Using the Quadratic Formula

Add 20u to both sides.

$$4u^{2}+25+20u=0$$

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.

$$4u^{2}+20u+25=0$$

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

$$u=\frac{-20±\sqrt{20^{2}-4\times 4\times 25}}{2\times 4}$$

Square 20.

$$u=\frac{-20±\sqrt{400-4\times 4\times 25}}{2\times 4}$$

Multiply -4 times 4.

$$u=\frac{-20±\sqrt{400-16\times 25}}{2\times 4}$$

Multiply -16 times 25.

$$u=\frac{-20±\sqrt{400-400}}{2\times 4}$$

Add 400 to -400.

$$u=\frac{-20±\sqrt{0}}{2\times 4}$$

Take the square root of 0.

$$u=-\frac{20}{2\times 4}$$

Multiply 2 times 4.

$$u=-\frac{20}{8}$$

Reduce the fraction \frac{-20}{8} to lowest terms by extracting and canceling out 4.

$$u=-\frac{5}{2}$$

Steps for Completing the Square

Add 20u to both sides.

$$4u^{2}+25+20u=0$$

Subtract 25 from both sides. Anything subtracted from zero gives its negation.

$$4u^{2}+20u=-25$$

Divide both sides by 4.

$$\frac{4u^{2}+20u}{4}=-\frac{25}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$u^{2}+\frac{20}{4}u=-\frac{25}{4}$$

Divide 20 by 4.

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

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.

$$u^{2}+5u+\left(\frac{5}{2}\right)^{2}=-\frac{25}{4}+\left(\frac{5}{2}\right)^{2}$$

Square \frac{5}{2} by squaring both the numerator and the denominator of the fraction.

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

Add -\frac{25}{4} to \frac{25}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$u^{2}+5u+\frac{25}{4}=0$$

Factor u^{2}+5u+\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}.

$$\left(u+\frac{5}{2}\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(u+\frac{5}{2}\right)^{2}}=\sqrt{0}$$

Simplify.

$$u+\frac{5}{2}=0 u+\frac{5}{2}=0$$

Subtract \frac{5}{2} from both sides of the equation.

$$u=-\frac{5}{2} u=-\frac{5}{2}$$

The equation is now solved. Solutions are the same.

$$u=-\frac{5}{2}$$