Solve p^2+22=10p-2

$$p ^ { 2 } + 22 = 10 p - 2$$

Solve for p

$$p=4$$

$$p=6$$

Steps Using Factoring

Subtract 10p from both sides.

$$p^{2}+22-10p=-2$$

Add 2 to both sides.

$$p^{2}+22-10p+2=0$$

Add 22 and 2 to get 24.

$$p^{2}+24-10p=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$p^{2}-10p+24=0$$

To solve the equation, factor p^{2}-10p+24 using formula p^{2}+\left(a+b\right)p+ab=\left(p+a\right)\left(p+b\right). To find a and b, set up a system to be solved.

$$a+b=-10 ab=24$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 24.

$$-1,-24 -2,-12 -3,-8 -4,-6$$

Calculate the sum for each pair.

$$-1-24=-25 -2-12=-14 -3-8=-11 -4-6=-10$$

The solution is the pair that gives sum -10.

$$a=-6 b=-4$$

Rewrite factored expression \left(p+a\right)\left(p+b\right) using the obtained values.

$$\left(p-6\right)\left(p-4\right)$$

To find equation solutions, solve p-6=0 and p-4=0.

$$p=6 p=4$$

Steps Using Factoring By Grouping

Subtract 10p from both sides.

$$p^{2}+22-10p=-2$$

Add 2 to both sides.

$$p^{2}+22-10p+2=0$$

Add 22 and 2 to get 24.

$$p^{2}+24-10p=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$p^{2}-10p+24=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as p^{2}+ap+bp+24. To find a and b, set up a system to be solved.

$$a+b=-10 ab=1\times 24=24$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 24.

$$-1,-24 -2,-12 -3,-8 -4,-6$$

Calculate the sum for each pair.

$$-1-24=-25 -2-12=-14 -3-8=-11 -4-6=-10$$

The solution is the pair that gives sum -10.

$$a=-6 b=-4$$

Rewrite p^{2}-10p+24 as \left(p^{2}-6p\right)+\left(-4p+24\right).

$$\left(p^{2}-6p\right)+\left(-4p+24\right)$$

Factor out p in the first and -4 in the second group.

$$p\left(p-6\right)-4\left(p-6\right)$$

Factor out common term p-6 by using distributive property.

$$\left(p-6\right)\left(p-4\right)$$

To find equation solutions, solve p-6=0 and p-4=0.

$$p=6 p=4$$

Steps Using the Quadratic Formula

Subtract 10p from both sides.

$$p^{2}+22-10p=-2$$

Add 2 to both sides.

$$p^{2}+22-10p+2=0$$

Add 22 and 2 to get 24.

$$p^{2}+24-10p=0$$

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.

$$p^{2}-10p+24=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -10 for b, and 24 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$p=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 24}}{2}$$

Square -10.

$$p=\frac{-\left(-10\right)±\sqrt{100-4\times 24}}{2}$$

Multiply -4 times 24.

$$p=\frac{-\left(-10\right)±\sqrt{100-96}}{2}$$

Add 100 to -96.

$$p=\frac{-\left(-10\right)±\sqrt{4}}{2}$$

Take the square root of 4.

$$p=\frac{-\left(-10\right)±2}{2}$$

The opposite of -10 is 10.

$$p=\frac{10±2}{2}$$

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

$$p=\frac{12}{2}$$

Divide 12 by 2.

$$p=6$$

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

$$p=\frac{8}{2}$$

Divide 8 by 2.

$$p=4$$

The equation is now solved.

$$p=6 p=4$$

Steps for Completing the Square

Subtract 10p from both sides.

$$p^{2}+22-10p=-2$$

Subtract 22 from both sides.

$$p^{2}-10p=-2-22$$

Subtract 22 from -2 to get -24.

$$p^{2}-10p=-24$$

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.

$$p^{2}-10p+\left(-5\right)^{2}=-24+\left(-5\right)^{2}$$

Square -5.

$$p^{2}-10p+25=-24+25$$

Add -24 to 25.

$$p^{2}-10p+25=1$$

Factor p^{2}-10p+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}.

$$\left(p-5\right)^{2}=1$$

Take the square root of both sides of the equation.

$$\sqrt{\left(p-5\right)^{2}}=\sqrt{1}$$

Simplify.

$$p-5=1 p-5=-1$$

Add 5 to both sides of the equation.

$$p=6 p=4$$