Solve 5x^2+96x-64*108=0

$$5 x ^ { 2 } + 96 x - 64 \times 108 = 0$$

Solve for x

$$x=-48$$

$$x = \frac{144}{5} = 28\frac{4}{5} = 28.8$$

Steps Using Factoring By Grouping

Multiply 64 and 108 to get 6912.

$$5x^{2}+96x-6912=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 5x^{2}+ax+bx-6912. To find a and b, set up a system to be solved.

$$a+b=96 ab=5\left(-6912\right)=-34560$$

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 -34560.

$$-1,34560 -2,17280 -3,11520 -4,8640 -5,6912 -6,5760 -8,4320 -9,3840 -10,3456 -12,2880 -15,2304 -16,2160 -18,1920 -20,1728 -24,1440 -27,1280 -30,1152 -32,1080 -36,960 -40,864 -45,768 -48,720 -54,640 -60,576 -64,540 -72,480 -80,432 -90,384 -96,360 -108,320 -120,288 -128,270 -135,256 -144,240 -160,216 -180,192$$

Calculate the sum for each pair.

$$-1+34560=34559 -2+17280=17278 -3+11520=11517 -4+8640=8636 -5+6912=6907 -6+5760=5754 -8+4320=4312 -9+3840=3831 -10+3456=3446 -12+2880=2868 -15+2304=2289 -16+2160=2144 -18+1920=1902 -20+1728=1708 -24+1440=1416 -27+1280=1253 -30+1152=1122 -32+1080=1048 -36+960=924 -40+864=824 -45+768=723 -48+720=672 -54+640=586 -60+576=516 -64+540=476 -72+480=408 -80+432=352 -90+384=294 -96+360=264 -108+320=212 -120+288=168 -128+270=142 -135+256=121 -144+240=96 -160+216=56 -180+192=12$$

The solution is the pair that gives sum 96.

$$a=-144 b=240$$

Rewrite 5x^{2}+96x-6912 as \left(5x^{2}-144x\right)+\left(240x-6912\right).

$$\left(5x^{2}-144x\right)+\left(240x-6912\right)$$

Factor out x in the first and 48 in the second group.

$$x\left(5x-144\right)+48\left(5x-144\right)$$

Factor out common term 5x-144 by using distributive property.

$$\left(5x-144\right)\left(x+48\right)$$

To find equation solutions, solve 5x-144=0 and x+48=0.

$$x=\frac{144}{5} x=-48$$

Steps Using the Quadratic Formula

Multiply 64 and 108 to get 6912.

$$5x^{2}+96x-6912=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, 96 for b, and -6912 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-96±\sqrt{96^{2}-4\times 5\left(-6912\right)}}{2\times 5}$$

Square 96.

$$x=\frac{-96±\sqrt{9216-4\times 5\left(-6912\right)}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-96±\sqrt{9216-20\left(-6912\right)}}{2\times 5}$$

Multiply -20 times -6912.

$$x=\frac{-96±\sqrt{9216+138240}}{2\times 5}$$

Add 9216 to 138240.

$$x=\frac{-96±\sqrt{147456}}{2\times 5}$$

Take the square root of 147456.

$$x=\frac{-96±384}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{-96±384}{10}$$

Now solve the equation x=\frac{-96±384}{10} when ± is plus. Add -96 to 384.

$$x=\frac{288}{10}$$

Reduce the fraction \frac{288}{10} to lowest terms by extracting and canceling out 2.

$$x=\frac{144}{5}$$

Now solve the equation x=\frac{-96±384}{10} when ± is minus. Subtract 384 from -96.

$$x=-\frac{480}{10}$$

Divide -480 by 10.

$$x=-48$$

The equation is now solved.

$$x=\frac{144}{5} x=-48$$

Steps for Completing the Square

Multiply 64 and 108 to get 6912.

$$5x^{2}+96x-6912=0$$

Add 6912 to both sides. Anything plus zero gives itself.

$$5x^{2}+96x=6912$$

Divide both sides by 5.

$$\frac{5x^{2}+96x}{5}=\frac{6912}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$x^{2}+\frac{96}{5}x=\frac{6912}{5}$$

Divide \frac{96}{5}, the coefficient of the x term, by 2 to get \frac{48}{5}. Then add the square of \frac{48}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{96}{5}x+\left(\frac{48}{5}\right)^{2}=\frac{6912}{5}+\left(\frac{48}{5}\right)^{2}$$

Square \frac{48}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{96}{5}x+\frac{2304}{25}=\frac{6912}{5}+\frac{2304}{25}$$

Add \frac{6912}{5} to \frac{2304}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{96}{5}x+\frac{2304}{25}=\frac{36864}{25}$$

Factor x^{2}+\frac{96}{5}x+\frac{2304}{25}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{48}{5}\right)^{2}=\frac{36864}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{48}{5}\right)^{2}}=\sqrt{\frac{36864}{25}}$$

Simplify.

$$x+\frac{48}{5}=\frac{192}{5} x+\frac{48}{5}=-\frac{192}{5}$$

Subtract \frac{48}{5} from both sides of the equation.

$$x=\frac{144}{5} x=-48$$