Solve 80x^2+160x-240=0

$$80 x ^ { 2 } + 160 x - 240 = 0$$

Solve for x

$$x=-3$$

$$x=1$$

Steps Using Factoring By Grouping

Divide both sides by 80.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-3. To find a and b, set up a system to be solved.

$$a+b=2 ab=1\left(-3\right)=-3$$

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. The only such pair is the system solution.

$$a=-1 b=3$$

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

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

Factor out x in the first and 3 in the second group.

$$x\left(x-1\right)+3\left(x-1\right)$$

Factor out common term x-1 by using distributive property.

$$\left(x-1\right)\left(x+3\right)$$

To find equation solutions, solve x-1=0 and x+3=0.

$$x=1 x=-3$$

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.

$$80x^{2}+160x-240=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 80 for a, 160 for b, and -240 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-160±\sqrt{160^{2}-4\times 80\left(-240\right)}}{2\times 80}$$

Square 160.

$$x=\frac{-160±\sqrt{25600-4\times 80\left(-240\right)}}{2\times 80}$$

Multiply -4 times 80.

$$x=\frac{-160±\sqrt{25600-320\left(-240\right)}}{2\times 80}$$

Multiply -320 times -240.

$$x=\frac{-160±\sqrt{25600+76800}}{2\times 80}$$

Add 25600 to 76800.

$$x=\frac{-160±\sqrt{102400}}{2\times 80}$$

Take the square root of 102400.

$$x=\frac{-160±320}{2\times 80}$$

Multiply 2 times 80.

$$x=\frac{-160±320}{160}$$

Now solve the equation x=\frac{-160±320}{160} when ± is plus. Add -160 to 320.

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

Divide 160 by 160.

$$x=1$$

Now solve the equation x=\frac{-160±320}{160} when ± is minus. Subtract 320 from -160.

$$x=-\frac{480}{160}$$

Divide -480 by 160.

$$x=-3$$

The equation is now solved.

$$x=1 x=-3$$

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.

$$80x^{2}+160x-240=0$$

Add 240 to both sides of the equation.

$$80x^{2}+160x-240-\left(-240\right)=-\left(-240\right)$$

Subtracting -240 from itself leaves 0.

$$80x^{2}+160x=-\left(-240\right)$$

Subtract -240 from 0.

$$80x^{2}+160x=240$$

Divide both sides by 80.

$$\frac{80x^{2}+160x}{80}=\frac{240}{80}$$

Dividing by 80 undoes the multiplication by 80.

$$x^{2}+\frac{160}{80}x=\frac{240}{80}$$

Divide 160 by 80.

$$x^{2}+2x=\frac{240}{80}$$

Divide 240 by 80.

$$x^{2}+2x=3$$

Divide 2, the coefficient of the x term, by 2 to get 1. Then add the square of 1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+2x+1^{2}=3+1^{2}$$

Square 1.

$$x^{2}+2x+1=3+1$$

Add 3 to 1.

$$x^{2}+2x+1=4$$

Factor x^{2}+2x+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+1\right)^{2}=4$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{4}$$

Simplify.

$$x+1=2 x+1=-2$$

Subtract 1 from both sides of the equation.

$$x=1 x=-3$$

Steps Using Direct Factoring Method

Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.This is achieved by dividing both sides of the equation by 80

$$x ^ 2 +2x -3 = 0$$

Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C

$$r + s = -2 rs = -3$$

Two numbers r and s sum up to -2 exactly when the average of the two numbers is \frac{1}{2}*-2 = -1. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = -1 - u s = -1 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -3

$$(-1 - u) (-1 + u) = -3$$

Simplify by expanding (a -b) (a + b) = a^2 – b^2

$$1 - u^2 = -3$$

Simplify the expression by subtracting 1 on both sides

$$-u^2 = -3-1 = -4$$

Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

$$u^2 = 4 u = \pm\sqrt{4} = \pm 2$$

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.

$$r =-1 - 2 = -3 s = -1 + 2 = 1$$