Solve p^2=1600-18p

$$p ^ { 2 } = 1600 - 18 p$$

Solve for p

$$p=-50$$

$$p=32$$

Steps Using Factoring

Subtract 1600 from both sides.

$$p^{2}-1600=-18p$$

Add 18p to both sides.

$$p^{2}-1600+18p=0$$

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

$$p^{2}+18p-1600=0$$

To solve the equation, factor p^{2}+18p-1600 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.

$$a+b=18 ab=-1600$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -1600.

$$-1,1600 -2,800 -4,400 -5,320 -8,200 -10,160 -16,100 -20,80 -25,64 -32,50 -40,40$$

Calculate the sum for each pair.

$$-1+1600=1599 -2+800=798 -4+400=396 -5+320=315 -8+200=192 -10+160=150 -16+100=84 -20+80=60 -25+64=39 -32+50=18 -40+40=0$$

The solution is the pair that gives sum 18.

$$a=-32 b=50$$

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

$$\left(p-32\right)\left(p+50\right)$$

To find equation solutions, solve p-32=0 and p+50=0.

$$p=32 p=-50$$

Steps Using Factoring By Grouping

Subtract 1600 from both sides.

$$p^{2}-1600=-18p$$

Add 18p to both sides.

$$p^{2}-1600+18p=0$$

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

$$p^{2}+18p-1600=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as p^{2}+ap+bp-1600. To find a and b, set up a system to be solved.

$$a+b=18 ab=1\left(-1600\right)=-1600$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -1600.

$$-1,1600 -2,800 -4,400 -5,320 -8,200 -10,160 -16,100 -20,80 -25,64 -32,50 -40,40$$

Calculate the sum for each pair.

$$-1+1600=1599 -2+800=798 -4+400=396 -5+320=315 -8+200=192 -10+160=150 -16+100=84 -20+80=60 -25+64=39 -32+50=18 -40+40=0$$

The solution is the pair that gives sum 18.

$$a=-32 b=50$$

Rewrite p^{2}+18p-1600 as \left(p^{2}-32p\right)+\left(50p-1600\right).

$$\left(p^{2}-32p\right)+\left(50p-1600\right)$$

Factor out p in the first and 50 in the second group.

$$p\left(p-32\right)+50\left(p-32\right)$$

Factor out common term p-32 by using distributive property.

$$\left(p-32\right)\left(p+50\right)$$

To find equation solutions, solve p-32=0 and p+50=0.

$$p=32 p=-50$$

Steps Using the Quadratic Formula

Subtract 1600 from both sides.

$$p^{2}-1600=-18p$$

Add 18p to both sides.

$$p^{2}-1600+18p=0$$

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.

$$p^{2}+18p-1600=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 18 for b, and -1600 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$p=\frac{-18±\sqrt{18^{2}-4\left(-1600\right)}}{2}$$

Square 18.

$$p=\frac{-18±\sqrt{324-4\left(-1600\right)}}{2}$$

Multiply -4 times -1600.

$$p=\frac{-18±\sqrt{324+6400}}{2}$$

Add 324 to 6400.

$$p=\frac{-18±\sqrt{6724}}{2}$$

Take the square root of 6724.

$$p=\frac{-18±82}{2}$$

Now solve the equation p=\frac{-18±82}{2} when ± is plus. Add -18 to 82.

$$p=\frac{64}{2}$$

Divide 64 by 2.

$$p=32$$

Now solve the equation p=\frac{-18±82}{2} when ± is minus. Subtract 82 from -18.

$$p=-\frac{100}{2}$$

Divide -100 by 2.

$$p=-50$$

The equation is now solved.

$$p=32 p=-50$$

Steps for Completing the Square

Add 18p to both sides.

$$p^{2}+18p=1600$$

Divide 18, the coefficient of the x term, by 2 to get 9. Then add the square of 9 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$p^{2}+18p+9^{2}=1600+9^{2}$$

Square 9.

$$p^{2}+18p+81=1600+81$$

Add 1600 to 81.

$$p^{2}+18p+81=1681$$

Factor p^{2}+18p+81. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(p+9\right)^{2}=1681$$

Take the square root of both sides of the equation.

$$\sqrt{\left(p+9\right)^{2}}=\sqrt{1681}$$

Simplify.

$$p+9=41 p+9=-41$$

Subtract 9 from both sides of the equation.

$$p=32 p=-50$$