Solve 128x^2-256x-72=0

$$128 x ^ { 2 } - 256 x - 72 = 0$$

Solve for x

$$x=-\frac{1}{4}=-0.25$$

$$x = \frac{9}{4} = 2\frac{1}{4} = 2.25$$

Steps Using Factoring By Grouping

Divide both sides by 8.

$$16x^{2}-32x-9=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 16x^{2}+ax+bx-9. To find a and b, set up a system to be solved.

$$a+b=-32 ab=16\left(-9\right)=-144$$

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 -144.

$$1,-144 2,-72 3,-48 4,-36 6,-24 8,-18 9,-16 12,-12$$

Calculate the sum for each pair.

$$1-144=-143 2-72=-70 3-48=-45 4-36=-32 6-24=-18 8-18=-10 9-16=-7 12-12=0$$

The solution is the pair that gives sum -32.

$$a=-36 b=4$$

Rewrite 16x^{2}-32x-9 as \left(16x^{2}-36x\right)+\left(4x-9\right).

$$\left(16x^{2}-36x\right)+\left(4x-9\right)$$

Factor out 4x in 16x^{2}-36x.

$$4x\left(4x-9\right)+4x-9$$

Factor out common term 4x-9 by using distributive property.

$$\left(4x-9\right)\left(4x+1\right)$$

To find equation solutions, solve 4x-9=0 and 4x+1=0.

$$x=\frac{9}{4} x=-\frac{1}{4}$$

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.

$$128x^{2}-256x-72=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 128 for a, -256 for b, and -72 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-256\right)±\sqrt{\left(-256\right)^{2}-4\times 128\left(-72\right)}}{2\times 128}$$

Square -256.

$$x=\frac{-\left(-256\right)±\sqrt{65536-4\times 128\left(-72\right)}}{2\times 128}$$

Multiply -4 times 128.

$$x=\frac{-\left(-256\right)±\sqrt{65536-512\left(-72\right)}}{2\times 128}$$

Multiply -512 times -72.

$$x=\frac{-\left(-256\right)±\sqrt{65536+36864}}{2\times 128}$$

Add 65536 to 36864.

$$x=\frac{-\left(-256\right)±\sqrt{102400}}{2\times 128}$$

Take the square root of 102400.

$$x=\frac{-\left(-256\right)±320}{2\times 128}$$

The opposite of -256 is 256.

$$x=\frac{256±320}{2\times 128}$$

Multiply 2 times 128.

$$x=\frac{256±320}{256}$$

Now solve the equation x=\frac{256±320}{256} when ± is plus. Add 256 to 320.

$$x=\frac{576}{256}$$

Reduce the fraction \frac{576}{256} to lowest terms by extracting and canceling out 64.

$$x=\frac{9}{4}$$

Now solve the equation x=\frac{256±320}{256} when ± is minus. Subtract 320 from 256.

$$x=-\frac{64}{256}$$

Reduce the fraction \frac{-64}{256} to lowest terms by extracting and canceling out 64.

$$x=-\frac{1}{4}$$

The equation is now solved.

$$x=\frac{9}{4} x=-\frac{1}{4}$$

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.

$$128x^{2}-256x-72=0$$

Add 72 to both sides of the equation.

$$128x^{2}-256x-72-\left(-72\right)=-\left(-72\right)$$

Subtracting -72 from itself leaves 0.

$$128x^{2}-256x=-\left(-72\right)$$

Subtract -72 from 0.

$$128x^{2}-256x=72$$

Divide both sides by 128.

$$\frac{128x^{2}-256x}{128}=\frac{72}{128}$$

Dividing by 128 undoes the multiplication by 128.

$$x^{2}+\left(-\frac{256}{128}\right)x=\frac{72}{128}$$

Divide -256 by 128.

$$x^{2}-2x=\frac{72}{128}$$

Reduce the fraction \frac{72}{128} to lowest terms by extracting and canceling out 8.

$$x^{2}-2x=\frac{9}{16}$$

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.

$$x^{2}-2x+1=\frac{9}{16}+1$$

Add \frac{9}{16} to 1.

$$x^{2}-2x+1=\frac{25}{16}$$

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}.

$$\left(x-1\right)^{2}=\frac{25}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-1\right)^{2}}=\sqrt{\frac{25}{16}}$$

Simplify.

$$x-1=\frac{5}{4} x-1=-\frac{5}{4}$$

Add 1 to both sides of the equation.

$$x=\frac{9}{4} x=-\frac{1}{4}$$

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 128

$$x ^ 2 -2x -\frac{9}{16} = 0$$

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

$$r + s = 2 rs = -\frac{9}{16}$$

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 = -\frac{9}{16}

$$(1 - u) (1 + u) = -\frac{9}{16}$$

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

$$1 - u^2 = -\frac{9}{16}$$

Simplify the expression by subtracting 1 on both sides

$$-u^2 = -\frac{9}{16}-1 = -\frac{25}{16}$$

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

$$u^2 = \frac{25}{16} u = \pm\sqrt{\frac{25}{16}} = \pm \frac{5}{4}$$

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

$$r =1 - \frac{5}{4} = -0.250 s = 1 + \frac{5}{4} = 2.250$$