Solve (10-p/2)^2=18p

$$( 10 - \frac { p } { 2 } ) ^ { 2 } = 18 p$$

Solve for p

$$p=12\sqrt{19}+56\approx 108.306787322$$

$$p=56-12\sqrt{19}\approx 3.693212678$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(10-\frac{p}{2}\right)^{2}.

$$100+20\left(-\frac{p}{2}\right)+\left(-\frac{p}{2}\right)^{2}=18p$$

Cancel out 2, the greatest common factor in 20 and 2.

$$100-10p+\left(-\frac{p}{2}\right)^{2}=18p$$

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

$$100-10p+\left(\frac{p}{2}\right)^{2}=18p$$

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

$$100-10p+\frac{p^{2}}{2^{2}}=18p$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 100-10p times \frac{2^{2}}{2^{2}}.

$$\frac{\left(100-10p\right)\times 2^{2}}{2^{2}}+\frac{p^{2}}{2^{2}}=18p$$

Since \frac{\left(100-10p\right)\times 2^{2}}{2^{2}} and \frac{p^{2}}{2^{2}} have the same denominator, add them by adding their numerators.

$$\frac{\left(100-10p\right)\times 2^{2}+p^{2}}{2^{2}}=18p$$

Do the multiplications in \left(100-10p\right)\times 2^{2}+p^{2}.

$$\frac{400-40p+p^{2}}{2^{2}}=18p$$

Subtract 18p from both sides.

$$\frac{400-40p+p^{2}}{2^{2}}-18p=0$$

Calculate 2 to the power of 2 and get 4.

$$\frac{400-40p+p^{2}}{4}-18p=0$$

Divide each term of 400-40p+p^{2} by 4 to get 100-10p+\frac{1}{4}p^{2}.

$$100-10p+\frac{1}{4}p^{2}-18p=0$$

Combine -10p and -18p to get -28p.

$$100-28p+\frac{1}{4}p^{2}=0$$

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.

$$\frac{1}{4}p^{2}-28p+100=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{1}{4} for a, -28 for b, and 100 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$p=\frac{-\left(-28\right)±\sqrt{\left(-28\right)^{2}-4\times \frac{1}{4}\times 100}}{2\times \frac{1}{4}}$$

Square -28.

$$p=\frac{-\left(-28\right)±\sqrt{784-4\times \frac{1}{4}\times 100}}{2\times \frac{1}{4}}$$

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

$$p=\frac{-\left(-28\right)±\sqrt{784-100}}{2\times \frac{1}{4}}$$

Add 784 to -100.

$$p=\frac{-\left(-28\right)±\sqrt{684}}{2\times \frac{1}{4}}$$

Take the square root of 684.

$$p=\frac{-\left(-28\right)±6\sqrt{19}}{2\times \frac{1}{4}}$$

The opposite of -28 is 28.

$$p=\frac{28±6\sqrt{19}}{2\times \frac{1}{4}}$$

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

$$p=\frac{28±6\sqrt{19}}{\frac{1}{2}}$$

Now solve the equation p=\frac{28±6\sqrt{19}}{\frac{1}{2}} when ± is plus. Add 28 to 6\sqrt{19}.

$$p=\frac{6\sqrt{19}+28}{\frac{1}{2}}$$

Divide 28+6\sqrt{19} by \frac{1}{2} by multiplying 28+6\sqrt{19} by the reciprocal of \frac{1}{2}.

$$p=12\sqrt{19}+56$$

Now solve the equation p=\frac{28±6\sqrt{19}}{\frac{1}{2}} when ± is minus. Subtract 6\sqrt{19} from 28.

$$p=\frac{28-6\sqrt{19}}{\frac{1}{2}}$$

Divide 28-6\sqrt{19} by \frac{1}{2} by multiplying 28-6\sqrt{19} by the reciprocal of \frac{1}{2}.

$$p=56-12\sqrt{19}$$

The equation is now solved.

$$p=12\sqrt{19}+56 p=56-12\sqrt{19}$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(10-\frac{p}{2}\right)^{2}.

$$100+20\left(-\frac{p}{2}\right)+\left(-\frac{p}{2}\right)^{2}=18p$$

Cancel out 2, the greatest common factor in 20 and 2.

$$100-10p+\left(-\frac{p}{2}\right)^{2}=18p$$

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

$$100-10p+\left(\frac{p}{2}\right)^{2}=18p$$

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

$$100-10p+\frac{p^{2}}{2^{2}}=18p$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 100-10p times \frac{2^{2}}{2^{2}}.

$$\frac{\left(100-10p\right)\times 2^{2}}{2^{2}}+\frac{p^{2}}{2^{2}}=18p$$

Since \frac{\left(100-10p\right)\times 2^{2}}{2^{2}} and \frac{p^{2}}{2^{2}} have the same denominator, add them by adding their numerators.

$$\frac{\left(100-10p\right)\times 2^{2}+p^{2}}{2^{2}}=18p$$

Do the multiplications in \left(100-10p\right)\times 2^{2}+p^{2}.

$$\frac{400-40p+p^{2}}{2^{2}}=18p$$

Subtract 18p from both sides.

$$\frac{400-40p+p^{2}}{2^{2}}-18p=0$$

Calculate 2 to the power of 2 and get 4.

$$\frac{400-40p+p^{2}}{4}-18p=0$$

Divide each term of 400-40p+p^{2} by 4 to get 100-10p+\frac{1}{4}p^{2}.

$$100-10p+\frac{1}{4}p^{2}-18p=0$$

Combine -10p and -18p to get -28p.

$$100-28p+\frac{1}{4}p^{2}=0$$

Subtract 100 from both sides. Anything subtracted from zero gives its negation.

$$-28p+\frac{1}{4}p^{2}=-100$$

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.

$$\frac{1}{4}p^{2}-28p=-100$$

Multiply both sides by 4.

$$\frac{\frac{1}{4}p^{2}-28p}{\frac{1}{4}}=-\frac{100}{\frac{1}{4}}$$

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

$$p^{2}+\left(-\frac{28}{\frac{1}{4}}\right)p=-\frac{100}{\frac{1}{4}}$$

Divide -28 by \frac{1}{4} by multiplying -28 by the reciprocal of \frac{1}{4}.

$$p^{2}-112p=-\frac{100}{\frac{1}{4}}$$

Divide -100 by \frac{1}{4} by multiplying -100 by the reciprocal of \frac{1}{4}.

$$p^{2}-112p=-400$$

Divide -112, the coefficient of the x term, by 2 to get -56. Then add the square of -56 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$p^{2}-112p+\left(-56\right)^{2}=-400+\left(-56\right)^{2}$$

Square -56.

$$p^{2}-112p+3136=-400+3136$$

Add -400 to 3136.

$$p^{2}-112p+3136=2736$$

Factor p^{2}-112p+3136. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(p-56\right)^{2}=2736$$

Take the square root of both sides of the equation.

$$\sqrt{\left(p-56\right)^{2}}=\sqrt{2736}$$

Simplify.

$$p-56=12\sqrt{19} p-56=-12\sqrt{19}$$

Add 56 to both sides of the equation.

$$p=12\sqrt{19}+56 p=56-12\sqrt{19}$$