Solve (4-p)^2=(5p-10)^2

$$( 4 - p ) ^ { 2 } = ( 5 p - 10 ) ^ { 2 }$$

Solve for p

$$p = \frac{7}{3} = 2\frac{1}{3} \approx 2.333333333$$

$$p = \frac{3}{2} = 1\frac{1}{2} = 1.5$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4-p\right)^{2}.

$$16-8p+p^{2}=\left(5p-10\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5p-10\right)^{2}.

$$16-8p+p^{2}=25p^{2}-100p+100$$

Subtract 25p^{2} from both sides.

$$16-8p+p^{2}-25p^{2}=-100p+100$$

Combine p^{2} and -25p^{2} to get -24p^{2}.

$$16-8p-24p^{2}=-100p+100$$

Add 100p to both sides.

$$16-8p-24p^{2}+100p=100$$

Combine -8p and 100p to get 92p.

$$16+92p-24p^{2}=100$$

Subtract 100 from both sides.

$$16+92p-24p^{2}-100=0$$

Subtract 100 from 16 to get -84.

$$-84+92p-24p^{2}=0$$

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.

$$-24p^{2}+92p-84=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -24 for a, 92 for b, and -84 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$p=\frac{-92±\sqrt{92^{2}-4\left(-24\right)\left(-84\right)}}{2\left(-24\right)}$$

Square 92.

$$p=\frac{-92±\sqrt{8464-4\left(-24\right)\left(-84\right)}}{2\left(-24\right)}$$

Multiply -4 times -24.

$$p=\frac{-92±\sqrt{8464+96\left(-84\right)}}{2\left(-24\right)}$$

Multiply 96 times -84.

$$p=\frac{-92±\sqrt{8464-8064}}{2\left(-24\right)}$$

Add 8464 to -8064.

$$p=\frac{-92±\sqrt{400}}{2\left(-24\right)}$$

Take the square root of 400.

$$p=\frac{-92±20}{2\left(-24\right)}$$

Multiply 2 times -24.

$$p=\frac{-92±20}{-48}$$

Now solve the equation p=\frac{-92±20}{-48} when ± is plus. Add -92 to 20.

$$p=-\frac{72}{-48}$$

Reduce the fraction \frac{-72}{-48} to lowest terms by extracting and canceling out 24.

$$p=\frac{3}{2}$$

Now solve the equation p=\frac{-92±20}{-48} when ± is minus. Subtract 20 from -92.

$$p=-\frac{112}{-48}$$

Reduce the fraction \frac{-112}{-48} to lowest terms by extracting and canceling out 16.

$$p=\frac{7}{3}$$

The equation is now solved.

$$p=\frac{3}{2} p=\frac{7}{3}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4-p\right)^{2}.

$$16-8p+p^{2}=\left(5p-10\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5p-10\right)^{2}.

$$16-8p+p^{2}=25p^{2}-100p+100$$

Subtract 25p^{2} from both sides.

$$16-8p+p^{2}-25p^{2}=-100p+100$$

Combine p^{2} and -25p^{2} to get -24p^{2}.

$$16-8p-24p^{2}=-100p+100$$

Add 100p to both sides.

$$16-8p-24p^{2}+100p=100$$

Combine -8p and 100p to get 92p.

$$16+92p-24p^{2}=100$$

Subtract 16 from both sides.

$$92p-24p^{2}=100-16$$

Subtract 16 from 100 to get 84.

$$92p-24p^{2}=84$$

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.

$$-24p^{2}+92p=84$$

Divide both sides by -24.

$$\frac{-24p^{2}+92p}{-24}=\frac{84}{-24}$$

Dividing by -24 undoes the multiplication by -24.

$$p^{2}+\frac{92}{-24}p=\frac{84}{-24}$$

Reduce the fraction \frac{92}{-24} to lowest terms by extracting and canceling out 4.

$$p^{2}-\frac{23}{6}p=\frac{84}{-24}$$

Reduce the fraction \frac{84}{-24} to lowest terms by extracting and canceling out 12.

$$p^{2}-\frac{23}{6}p=-\frac{7}{2}$$

Divide -\frac{23}{6}, the coefficient of the x term, by 2 to get -\frac{23}{12}. Then add the square of -\frac{23}{12} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$p^{2}-\frac{23}{6}p+\left(-\frac{23}{12}\right)^{2}=-\frac{7}{2}+\left(-\frac{23}{12}\right)^{2}$$

Square -\frac{23}{12} by squaring both the numerator and the denominator of the fraction.

$$p^{2}-\frac{23}{6}p+\frac{529}{144}=-\frac{7}{2}+\frac{529}{144}$$

Add -\frac{7}{2} to \frac{529}{144} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$p^{2}-\frac{23}{6}p+\frac{529}{144}=\frac{25}{144}$$

Factor p^{2}-\frac{23}{6}p+\frac{529}{144}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(p-\frac{23}{12}\right)^{2}=\frac{25}{144}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(p-\frac{23}{12}\right)^{2}}=\sqrt{\frac{25}{144}}$$

Simplify.

$$p-\frac{23}{12}=\frac{5}{12} p-\frac{23}{12}=-\frac{5}{12}$$

Add \frac{23}{12} to both sides of the equation.

$$p=\frac{7}{3} p=\frac{3}{2}$$