Solve 100=-4b^2-40b+400

$$100 = - 4 b ^ { 2 } - 40 b + 400$$

Solve for b

$$b=-15$$

$$b=5$$

Steps Using Factoring By Grouping

Swap sides so that all variable terms are on the left hand side.

$$-4b^{2}-40b+400=100$$

Subtract 100 from both sides.

$$-4b^{2}-40b+400-100=0$$

Subtract 100 from 400 to get 300.

$$-4b^{2}-40b+300=0$$

Divide both sides by 4.

$$-b^{2}-10b+75=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -b^{2}+ab+bb+75. To find a and b, set up a system to be solved.

$$a+b=-10 ab=-75=-75$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -75.

$$1,-75 3,-25 5,-15$$

Calculate the sum for each pair.

$$1-75=-74 3-25=-22 5-15=-10$$

The solution is the pair that gives sum -10.

$$a=5 b=-15$$

Rewrite -b^{2}-10b+75 as \left(-b^{2}+5b\right)+\left(-15b+75\right).

$$\left(-b^{2}+5b\right)+\left(-15b+75\right)$$

Factor out b in the first and 15 in the second group.

$$b\left(-b+5\right)+15\left(-b+5\right)$$

Factor out common term -b+5 by using distributive property.

$$\left(-b+5\right)\left(b+15\right)$$

To find equation solutions, solve -b+5=0 and b+15=0.

$$b=5 b=-15$$

Steps Using the Quadratic Formula

Swap sides so that all variable terms are on the left hand side.

$$-4b^{2}-40b+400=100$$

Subtract 100 from both sides.

$$-4b^{2}-40b+400-100=0$$

Subtract 100 from 400 to get 300.

$$-4b^{2}-40b+300=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -4 for a, -40 for b, and 300 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$b=\frac{-\left(-40\right)±\sqrt{\left(-40\right)^{2}-4\left(-4\right)\times 300}}{2\left(-4\right)}$$

Square -40.

$$b=\frac{-\left(-40\right)±\sqrt{1600-4\left(-4\right)\times 300}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$b=\frac{-\left(-40\right)±\sqrt{1600+16\times 300}}{2\left(-4\right)}$$

Multiply 16 times 300.

$$b=\frac{-\left(-40\right)±\sqrt{1600+4800}}{2\left(-4\right)}$$

Add 1600 to 4800.

$$b=\frac{-\left(-40\right)±\sqrt{6400}}{2\left(-4\right)}$$

Take the square root of 6400.

$$b=\frac{-\left(-40\right)±80}{2\left(-4\right)}$$

The opposite of -40 is 40.

$$b=\frac{40±80}{2\left(-4\right)}$$

Multiply 2 times -4.

$$b=\frac{40±80}{-8}$$

Now solve the equation b=\frac{40±80}{-8} when ± is plus. Add 40 to 80.

$$b=\frac{120}{-8}$$

Divide 120 by -8.

$$b=-15$$

Now solve the equation b=\frac{40±80}{-8} when ± is minus. Subtract 80 from 40.

$$b=-\frac{40}{-8}$$

Divide -40 by -8.

$$b=5$$

The equation is now solved.

$$b=-15 b=5$$

Steps for Completing the Square

Swap sides so that all variable terms are on the left hand side.

$$-4b^{2}-40b+400=100$$

Subtract 400 from both sides.

$$-4b^{2}-40b=100-400$$

Subtract 400 from 100 to get -300.

$$-4b^{2}-40b=-300$$

Divide both sides by -4.

$$\frac{-4b^{2}-40b}{-4}=-\frac{300}{-4}$$

Dividing by -4 undoes the multiplication by -4.

$$b^{2}+\left(-\frac{40}{-4}\right)b=-\frac{300}{-4}$$

Divide -40 by -4.

$$b^{2}+10b=-\frac{300}{-4}$$

Divide -300 by -4.

$$b^{2}+10b=75$$

Divide 10, the coefficient of the x term, by 2 to get 5. Then add the square of 5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$b^{2}+10b+5^{2}=75+5^{2}$$

Square 5.

$$b^{2}+10b+25=75+25$$

Add 75 to 25.

$$b^{2}+10b+25=100$$

Factor b^{2}+10b+25. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(b+5\right)^{2}=100$$

Take the square root of both sides of the equation.

$$\sqrt{\left(b+5\right)^{2}}=\sqrt{100}$$

Simplify.

$$b+5=10 b+5=-10$$

Subtract 5 from both sides of the equation.

$$b=5 b=-15$$