Solve x^2-28x+160=0

$${ x }^{ 2 } -28x+160=0$$

Solve for x

$$x=8$$

$$x=20$$

Steps Using Factoring

To solve the equation, factor x^{2}-28x+160 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=-28 ab=160$$

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 160.

$$-1,-160 -2,-80 -4,-40 -5,-32 -8,-20 -10,-16$$

Calculate the sum for each pair.

$$-1-160=-161 -2-80=-82 -4-40=-44 -5-32=-37 -8-20=-28 -10-16=-26$$

The solution is the pair that gives sum -28.

$$a=-20 b=-8$$

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

$$\left(x-20\right)\left(x-8\right)$$

To find equation solutions, solve x-20=0 and x-8=0.

$$x=20 x=8$$

Steps Using Factoring By Grouping

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+160. To find a and b, set up a system to be solved.

$$a+b=-28 ab=1\times 160=160$$

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 160.

$$-1,-160 -2,-80 -4,-40 -5,-32 -8,-20 -10,-16$$

Calculate the sum for each pair.

$$-1-160=-161 -2-80=-82 -4-40=-44 -5-32=-37 -8-20=-28 -10-16=-26$$

The solution is the pair that gives sum -28.

$$a=-20 b=-8$$

Rewrite x^{2}-28x+160 as \left(x^{2}-20x\right)+\left(-8x+160\right).

$$\left(x^{2}-20x\right)+\left(-8x+160\right)$$

Factor out x in the first and -8 in the second group.

$$x\left(x-20\right)-8\left(x-20\right)$$

Factor out common term x-20 by using distributive property.

$$\left(x-20\right)\left(x-8\right)$$

To find equation solutions, solve x-20=0 and x-8=0.

$$x=20 x=8$$

Steps Using the Quadratic Formula

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.

$$x^{2}-28x+160=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -28 for b, and 160 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-28\right)±\sqrt{\left(-28\right)^{2}-4\times 160}}{2}$$

Square -28.

$$x=\frac{-\left(-28\right)±\sqrt{784-4\times 160}}{2}$$

Multiply -4 times 160.

$$x=\frac{-\left(-28\right)±\sqrt{784-640}}{2}$$

Add 784 to -640.

$$x=\frac{-\left(-28\right)±\sqrt{144}}{2}$$

Take the square root of 144.

$$x=\frac{-\left(-28\right)±12}{2}$$

The opposite of -28 is 28.

$$x=\frac{28±12}{2}$$

Now solve the equation x=\frac{28±12}{2} when ± is plus. Add 28 to 12.

$$x=\frac{40}{2}$$

Divide 40 by 2.

$$x=20$$

Now solve the equation x=\frac{28±12}{2} when ± is minus. Subtract 12 from 28.

$$x=\frac{16}{2}$$

Divide 16 by 2.

$$x=8$$

The equation is now solved.

$$x=20 x=8$$

Steps for Completing the Square

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.

$$x^{2}-28x+160=0$$

Subtract 160 from both sides of the equation.

$$x^{2}-28x+160-160=-160$$

Subtracting 160 from itself leaves 0.

$$x^{2}-28x=-160$$

Divide -28, the coefficient of the x term, by 2 to get -14. Then add the square of -14 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-28x+\left(-14\right)^{2}=-160+\left(-14\right)^{2}$$

Square -14.

$$x^{2}-28x+196=-160+196$$

Add -160 to 196.

$$x^{2}-28x+196=36$$

Factor x^{2}-28x+196. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-14\right)^{2}=36$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-14\right)^{2}}=\sqrt{36}$$

Simplify.

$$x-14=6 x-14=-6$$

Add 14 to both sides of the equation.

$$x=20 x=8$$