Solve 16x^2-48x-160=0

$$16 { x }^{ 2 } -48x-160=0$$

Solve for x

$$x=-2$$

$$x=5$$

Steps Using Factoring By Grouping

Divide both sides by 16.

$$x^{2}-3x-10=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-10. To find a and b, set up a system to be solved.

$$a+b=-3 ab=1\left(-10\right)=-10$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -10.

$$1,-10 2,-5$$

Calculate the sum for each pair.

$$1-10=-9 2-5=-3$$

The solution is the pair that gives sum -3.

$$a=-5 b=2$$

Rewrite x^{2}-3x-10 as \left(x^{2}-5x\right)+\left(2x-10\right).

$$\left(x^{2}-5x\right)+\left(2x-10\right)$$

Factor out x in the first and 2 in the second group.

$$x\left(x-5\right)+2\left(x-5\right)$$

Factor out common term x-5 by using distributive property.

$$\left(x-5\right)\left(x+2\right)$$

To find equation solutions, solve x-5=0 and x+2=0.

$$x=5 x=-2$$

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.

$$16x^{2}-48x-160=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 16 for a, -48 for b, and -160 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-48\right)±\sqrt{\left(-48\right)^{2}-4\times 16\left(-160\right)}}{2\times 16}$$

Square -48.

$$x=\frac{-\left(-48\right)±\sqrt{2304-4\times 16\left(-160\right)}}{2\times 16}$$

Multiply -4 times 16.

$$x=\frac{-\left(-48\right)±\sqrt{2304-64\left(-160\right)}}{2\times 16}$$

Multiply -64 times -160.

$$x=\frac{-\left(-48\right)±\sqrt{2304+10240}}{2\times 16}$$

Add 2304 to 10240.

$$x=\frac{-\left(-48\right)±\sqrt{12544}}{2\times 16}$$

Take the square root of 12544.

$$x=\frac{-\left(-48\right)±112}{2\times 16}$$

The opposite of -48 is 48.

$$x=\frac{48±112}{2\times 16}$$

Multiply 2 times 16.

$$x=\frac{48±112}{32}$$

Now solve the equation x=\frac{48±112}{32} when ± is plus. Add 48 to 112.

$$x=\frac{160}{32}$$

Divide 160 by 32.

$$x=5$$

Now solve the equation x=\frac{48±112}{32} when ± is minus. Subtract 112 from 48.

$$x=-\frac{64}{32}$$

Divide -64 by 32.

$$x=-2$$

The equation is now solved.

$$x=5 x=-2$$

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.

$$16x^{2}-48x-160=0$$

Add 160 to both sides of the equation.

$$16x^{2}-48x-160-\left(-160\right)=-\left(-160\right)$$

Subtracting -160 from itself leaves 0.

$$16x^{2}-48x=-\left(-160\right)$$

Subtract -160 from 0.

$$16x^{2}-48x=160$$

Divide both sides by 16.

$$\frac{16x^{2}-48x}{16}=\frac{160}{16}$$

Dividing by 16 undoes the multiplication by 16.

$$x^{2}+\left(-\frac{48}{16}\right)x=\frac{160}{16}$$

Divide -48 by 16.

$$x^{2}-3x=\frac{160}{16}$$

Divide 160 by 16.

$$x^{2}-3x=10$$

Divide -3, the coefficient of the x term, by 2 to get -\frac{3}{2}. Then add the square of -\frac{3}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-3x+\left(-\frac{3}{2}\right)^{2}=10+\left(-\frac{3}{2}\right)^{2}$$

Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-3x+\frac{9}{4}=10+\frac{9}{4}$$

Add 10 to \frac{9}{4}.

$$x^{2}-3x+\frac{9}{4}=\frac{49}{4}$$

Factor x^{2}-3x+\frac{9}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{3}{2}\right)^{2}=\frac{49}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{2}\right)^{2}}=\sqrt{\frac{49}{4}}$$

Simplify.

$$x-\frac{3}{2}=\frac{7}{2} x-\frac{3}{2}=-\frac{7}{2}$$

Add \frac{3}{2} to both sides of the equation.

$$x=5 x=-2$$