Solve (25-3b)-(-1/4b^2+2b)=16

$$( 25 - 3 b ) - ( - \frac { 1 } { 4 } b ^ { 2 } + 2 b ) = 16$$

Solve for b

$$b=2$$

$$b=18$$

Steps Using the Quadratic Formula

To find the opposite of -\frac{1}{4}b^{2}+2b, find the opposite of each term.

$$25-3b+\frac{1}{4}b^{2}-2b=16$$

Combine -3b and -2b to get -5b.

$$25-5b+\frac{1}{4}b^{2}=16$$

Subtract 16 from both sides.

$$25-5b+\frac{1}{4}b^{2}-16=0$$

Subtract 16 from 25 to get 9.

$$9-5b+\frac{1}{4}b^{2}=0$$

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.

$$\frac{1}{4}b^{2}-5b+9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{1}{4} for a, -5 for b, and 9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$b=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times \frac{1}{4}\times 9}}{2\times \frac{1}{4}}$$

Square -5.

$$b=\frac{-\left(-5\right)±\sqrt{25-4\times \frac{1}{4}\times 9}}{2\times \frac{1}{4}}$$

Multiply -4 times \frac{1}{4}.

$$b=\frac{-\left(-5\right)±\sqrt{25-9}}{2\times \frac{1}{4}}$$

Add 25 to -9.

$$b=\frac{-\left(-5\right)±\sqrt{16}}{2\times \frac{1}{4}}$$

Take the square root of 16.

$$b=\frac{-\left(-5\right)±4}{2\times \frac{1}{4}}$$

The opposite of -5 is 5.

$$b=\frac{5±4}{2\times \frac{1}{4}}$$

Multiply 2 times \frac{1}{4}.

$$b=\frac{5±4}{\frac{1}{2}}$$

Now solve the equation b=\frac{5±4}{\frac{1}{2}} when ± is plus. Add 5 to 4.

$$b=\frac{9}{\frac{1}{2}}$$

Divide 9 by \frac{1}{2} by multiplying 9 by the reciprocal of \frac{1}{2}.

$$b=18$$

Now solve the equation b=\frac{5±4}{\frac{1}{2}} when ± is minus. Subtract 4 from 5.

$$b=\frac{1}{\frac{1}{2}}$$

Divide 1 by \frac{1}{2} by multiplying 1 by the reciprocal of \frac{1}{2}.

$$b=2$$

The equation is now solved.

$$b=18 b=2$$

Steps for Completing the Square

To find the opposite of -\frac{1}{4}b^{2}+2b, find the opposite of each term.

$$25-3b+\frac{1}{4}b^{2}-2b=16$$

Combine -3b and -2b to get -5b.

$$25-5b+\frac{1}{4}b^{2}=16$$

Subtract 25 from both sides.

$$-5b+\frac{1}{4}b^{2}=16-25$$

Subtract 25 from 16 to get -9.

$$-5b+\frac{1}{4}b^{2}=-9$$

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.

$$\frac{1}{4}b^{2}-5b=-9$$

Multiply both sides by 4.

$$\frac{\frac{1}{4}b^{2}-5b}{\frac{1}{4}}=-\frac{9}{\frac{1}{4}}$$

Dividing by \frac{1}{4} undoes the multiplication by \frac{1}{4}.

$$b^{2}+\left(-\frac{5}{\frac{1}{4}}\right)b=-\frac{9}{\frac{1}{4}}$$

Divide -5 by \frac{1}{4} by multiplying -5 by the reciprocal of \frac{1}{4}.

$$b^{2}-20b=-\frac{9}{\frac{1}{4}}$$

Divide -9 by \frac{1}{4} by multiplying -9 by the reciprocal of \frac{1}{4}.

$$b^{2}-20b=-36$$

Divide -20, the coefficient of the x term, by 2 to get -10. Then add the square of -10 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$b^{2}-20b+\left(-10\right)^{2}=-36+\left(-10\right)^{2}$$

Square -10.

$$b^{2}-20b+100=-36+100$$

Add -36 to 100.

$$b^{2}-20b+100=64$$

Factor b^{2}-20b+100. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(b-10\right)^{2}=64$$

Take the square root of both sides of the equation.

$$\sqrt{\left(b-10\right)^{2}}=\sqrt{64}$$

Simplify.

$$b-10=8 b-10=-8$$

Add 10 to both sides of the equation.

$$b=18 b=2$$