Solve 18p+(9-p/2)^2=100

$$18 p + ( 9 - \frac { p } { 2 } ) ^ { 2 } = 100$$

Solve for p

$$p=-38$$

$$p=2$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(9-\frac{p}{2}\right)^{2}.

$$18p+81+18\left(-\frac{p}{2}\right)+\left(-\frac{p}{2}\right)^{2}=100$$

Cancel out 2, the greatest common factor in 18 and 2.

$$18p+81-9p+\left(-\frac{p}{2}\right)^{2}=100$$

Calculate -\frac{p}{2} to the power of 2 and get \left(\frac{p}{2}\right)^{2}.

$$18p+81-9p+\left(\frac{p}{2}\right)^{2}=100$$

Combine 18p and -9p to get 9p.

$$9p+81+\left(\frac{p}{2}\right)^{2}=100$$

To raise \frac{p}{2} to a power, raise both numerator and denominator to the power and then divide.

$$9p+81+\frac{p^{2}}{2^{2}}=100$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 9p+81 times \frac{2^{2}}{2^{2}}.

$$\frac{\left(9p+81\right)\times 2^{2}}{2^{2}}+\frac{p^{2}}{2^{2}}=100$$

Since \frac{\left(9p+81\right)\times 2^{2}}{2^{2}} and \frac{p^{2}}{2^{2}} have the same denominator, add them by adding their numerators.

$$\frac{\left(9p+81\right)\times 2^{2}+p^{2}}{2^{2}}=100$$

Do the multiplications in \left(9p+81\right)\times 2^{2}+p^{2}.

$$\frac{36p+324+p^{2}}{2^{2}}=100$$

Calculate 2 to the power of 2 and get 4.

$$\frac{36p+324+p^{2}}{4}=100$$

Divide each term of 36p+324+p^{2} by 4 to get 9p+81+\frac{1}{4}p^{2}.

$$9p+81+\frac{1}{4}p^{2}=100$$

Subtract 100 from both sides.

$$9p+81+\frac{1}{4}p^{2}-100=0$$

Subtract 100 from 81 to get -19.

$$9p-19+\frac{1}{4}p^{2}=0$$

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.

$$\frac{1}{4}p^{2}+9p-19=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{1}{4} for a, 9 for b, and -19 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$p=\frac{-9±\sqrt{9^{2}-4\times \frac{1}{4}\left(-19\right)}}{2\times \frac{1}{4}}$$

Square 9.

$$p=\frac{-9±\sqrt{81-4\times \frac{1}{4}\left(-19\right)}}{2\times \frac{1}{4}}$$

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

$$p=\frac{-9±\sqrt{81-\left(-19\right)}}{2\times \frac{1}{4}}$$

Multiply -1 times -19.

$$p=\frac{-9±\sqrt{81+19}}{2\times \frac{1}{4}}$$

Add 81 to 19.

$$p=\frac{-9±\sqrt{100}}{2\times \frac{1}{4}}$$

Take the square root of 100.

$$p=\frac{-9±10}{2\times \frac{1}{4}}$$

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

$$p=\frac{-9±10}{\frac{1}{2}}$$

Now solve the equation p=\frac{-9±10}{\frac{1}{2}} when ± is plus. Add -9 to 10.

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

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

$$p=2$$

Now solve the equation p=\frac{-9±10}{\frac{1}{2}} when ± is minus. Subtract 10 from -9.

$$p=-\frac{19}{\frac{1}{2}}$$

Divide -19 by \frac{1}{2} by multiplying -19 by the reciprocal of \frac{1}{2}.

$$p=-38$$

The equation is now solved.

$$p=2 p=-38$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(9-\frac{p}{2}\right)^{2}.

$$18p+81+18\left(-\frac{p}{2}\right)+\left(-\frac{p}{2}\right)^{2}=100$$

Cancel out 2, the greatest common factor in 18 and 2.

$$18p+81-9p+\left(-\frac{p}{2}\right)^{2}=100$$

Calculate -\frac{p}{2} to the power of 2 and get \left(\frac{p}{2}\right)^{2}.

$$18p+81-9p+\left(\frac{p}{2}\right)^{2}=100$$

Combine 18p and -9p to get 9p.

$$9p+81+\left(\frac{p}{2}\right)^{2}=100$$

To raise \frac{p}{2} to a power, raise both numerator and denominator to the power and then divide.

$$9p+81+\frac{p^{2}}{2^{2}}=100$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 9p+81 times \frac{2^{2}}{2^{2}}.

$$\frac{\left(9p+81\right)\times 2^{2}}{2^{2}}+\frac{p^{2}}{2^{2}}=100$$

Since \frac{\left(9p+81\right)\times 2^{2}}{2^{2}} and \frac{p^{2}}{2^{2}} have the same denominator, add them by adding their numerators.

$$\frac{\left(9p+81\right)\times 2^{2}+p^{2}}{2^{2}}=100$$

Do the multiplications in \left(9p+81\right)\times 2^{2}+p^{2}.

$$\frac{36p+324+p^{2}}{2^{2}}=100$$

Calculate 2 to the power of 2 and get 4.

$$\frac{36p+324+p^{2}}{4}=100$$

Divide each term of 36p+324+p^{2} by 4 to get 9p+81+\frac{1}{4}p^{2}.

$$9p+81+\frac{1}{4}p^{2}=100$$

Subtract 81 from both sides.

$$9p+\frac{1}{4}p^{2}=100-81$$

Subtract 81 from 100 to get 19.

$$9p+\frac{1}{4}p^{2}=19$$

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.

$$\frac{1}{4}p^{2}+9p=19$$

Multiply both sides by 4.

$$\frac{\frac{1}{4}p^{2}+9p}{\frac{1}{4}}=\frac{19}{\frac{1}{4}}$$

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

$$p^{2}+\frac{9}{\frac{1}{4}}p=\frac{19}{\frac{1}{4}}$$

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

$$p^{2}+36p=\frac{19}{\frac{1}{4}}$$

Divide 19 by \frac{1}{4} by multiplying 19 by the reciprocal of \frac{1}{4}.

$$p^{2}+36p=76$$

Divide 36, the coefficient of the x term, by 2 to get 18. Then add the square of 18 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$p^{2}+36p+18^{2}=76+18^{2}$$

Square 18.

$$p^{2}+36p+324=76+324$$

Add 76 to 324.

$$p^{2}+36p+324=400$$

Factor p^{2}+36p+324. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(p+18\right)^{2}=400$$

Take the square root of both sides of the equation.

$$\sqrt{\left(p+18\right)^{2}}=\sqrt{400}$$

Simplify.

$$p+18=20 p+18=-20$$

Subtract 18 from both sides of the equation.

$$p=2 p=-38$$