Solve 5x-4-(12x+6)(5x-4)/frac{1{2}}=0

$$5 x - 4 - \frac { ( 12 x + 6 ) ( 5 x - 4 ) } { \frac { 1 } { 2 } } = 0$$

Solve for x

$$x=\frac{4}{5}=0.8$$

$$x=-\frac{11}{24}\approx -0.458333333$$

Steps Using Factoring By Grouping

Use the distributive property to multiply 12x+6 by 5x-4 and combine like terms.

$$5x-4-\frac{60x^{2}-18x-24}{\frac{1}{2}}=0$$

Divide each term of 60x^{2}-18x-24 by \frac{1}{2} to get \frac{60x^{2}}{\frac{1}{2}}+\frac{-18x}{\frac{1}{2}}+\frac{-24}{\frac{1}{2}}.

$$5x-4-\left(\frac{60x^{2}}{\frac{1}{2}}+\frac{-18x}{\frac{1}{2}}+\frac{-24}{\frac{1}{2}}\right)=0$$

Divide 60x^{2} by \frac{1}{2} to get 120x^{2}.

$$5x-4-\left(120x^{2}+\frac{-18x}{\frac{1}{2}}+\frac{-24}{\frac{1}{2}}\right)=0$$

Divide -18x by \frac{1}{2} to get -36x.

$$5x-4-\left(120x^{2}-36x+\frac{-24}{\frac{1}{2}}\right)=0$$

Divide -24 by \frac{1}{2} by multiplying -24 by the reciprocal of \frac{1}{2}.

$$5x-4-\left(120x^{2}-36x-24\times 2\right)=0$$

Multiply -24 and 2 to get -48.

$$5x-4-\left(120x^{2}-36x-48\right)=0$$

To find the opposite of 120x^{2}-36x-48, find the opposite of each term.

$$5x-4-120x^{2}+36x+48=0$$

Combine 5x and 36x to get 41x.

$$41x-4-120x^{2}+48=0$$

Add -4 and 48 to get 44.

$$41x+44-120x^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-120x^{2}+41x+44=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -120x^{2}+ax+bx+44. To find a and b, set up a system to be solved.

$$a+b=41 ab=-120\times 44=-5280$$

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 -5280.

$$-1,5280 -2,2640 -3,1760 -4,1320 -5,1056 -6,880 -8,660 -10,528 -11,480 -12,440 -15,352 -16,330 -20,264 -22,240 -24,220 -30,176 -32,165 -33,160 -40,132 -44,120 -48,110 -55,96 -60,88 -66,80$$

Calculate the sum for each pair.

$$-1+5280=5279 -2+2640=2638 -3+1760=1757 -4+1320=1316 -5+1056=1051 -6+880=874 -8+660=652 -10+528=518 -11+480=469 -12+440=428 -15+352=337 -16+330=314 -20+264=244 -22+240=218 -24+220=196 -30+176=146 -32+165=133 -33+160=127 -40+132=92 -44+120=76 -48+110=62 -55+96=41 -60+88=28 -66+80=14$$

The solution is the pair that gives sum 41.

$$a=96 b=-55$$

Rewrite -120x^{2}+41x+44 as \left(-120x^{2}+96x\right)+\left(-55x+44\right).

$$\left(-120x^{2}+96x\right)+\left(-55x+44\right)$$

Factor out -24x in the first and -11 in the second group.

$$-24x\left(5x-4\right)-11\left(5x-4\right)$$

Factor out common term 5x-4 by using distributive property.

$$\left(5x-4\right)\left(-24x-11\right)$$

To find equation solutions, solve 5x-4=0 and -24x-11=0.

$$x=\frac{4}{5} x=-\frac{11}{24}$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 12x+6 by 5x-4 and combine like terms.

$$5x-4-\frac{60x^{2}-18x-24}{\frac{1}{2}}=0$$

Divide each term of 60x^{2}-18x-24 by \frac{1}{2} to get \frac{60x^{2}}{\frac{1}{2}}+\frac{-18x}{\frac{1}{2}}+\frac{-24}{\frac{1}{2}}.

$$5x-4-\left(\frac{60x^{2}}{\frac{1}{2}}+\frac{-18x}{\frac{1}{2}}+\frac{-24}{\frac{1}{2}}\right)=0$$

Divide 60x^{2} by \frac{1}{2} to get 120x^{2}.

$$5x-4-\left(120x^{2}+\frac{-18x}{\frac{1}{2}}+\frac{-24}{\frac{1}{2}}\right)=0$$

Divide -18x by \frac{1}{2} to get -36x.

$$5x-4-\left(120x^{2}-36x+\frac{-24}{\frac{1}{2}}\right)=0$$

Divide -24 by \frac{1}{2} by multiplying -24 by the reciprocal of \frac{1}{2}.

$$5x-4-\left(120x^{2}-36x-24\times 2\right)=0$$

Multiply -24 and 2 to get -48.

$$5x-4-\left(120x^{2}-36x-48\right)=0$$

To find the opposite of 120x^{2}-36x-48, find the opposite of each term.

$$5x-4-120x^{2}+36x+48=0$$

Combine 5x and 36x to get 41x.

$$41x-4-120x^{2}+48=0$$

Add -4 and 48 to get 44.

$$41x+44-120x^{2}=0$$

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.

$$-120x^{2}+41x+44=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -120 for a, 41 for b, and 44 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-41±\sqrt{41^{2}-4\left(-120\right)\times 44}}{2\left(-120\right)}$$

Square 41.

$$x=\frac{-41±\sqrt{1681-4\left(-120\right)\times 44}}{2\left(-120\right)}$$

Multiply -4 times -120.

$$x=\frac{-41±\sqrt{1681+480\times 44}}{2\left(-120\right)}$$

Multiply 480 times 44.

$$x=\frac{-41±\sqrt{1681+21120}}{2\left(-120\right)}$$

Add 1681 to 21120.

$$x=\frac{-41±\sqrt{22801}}{2\left(-120\right)}$$

Take the square root of 22801.

$$x=\frac{-41±151}{2\left(-120\right)}$$

Multiply 2 times -120.

$$x=\frac{-41±151}{-240}$$

Now solve the equation x=\frac{-41±151}{-240} when ± is plus. Add -41 to 151.

$$x=\frac{110}{-240}$$

Reduce the fraction \frac{110}{-240} to lowest terms by extracting and canceling out 10.

$$x=-\frac{11}{24}$$

Now solve the equation x=\frac{-41±151}{-240} when ± is minus. Subtract 151 from -41.

$$x=-\frac{192}{-240}$$

Reduce the fraction \frac{-192}{-240} to lowest terms by extracting and canceling out 48.

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

The equation is now solved.

$$x=-\frac{11}{24} x=\frac{4}{5}$$

Steps for Completing the Square

Use the distributive property to multiply 12x+6 by 5x-4 and combine like terms.

$$5x-4-\frac{60x^{2}-18x-24}{\frac{1}{2}}=0$$

Divide each term of 60x^{2}-18x-24 by \frac{1}{2} to get \frac{60x^{2}}{\frac{1}{2}}+\frac{-18x}{\frac{1}{2}}+\frac{-24}{\frac{1}{2}}.

$$5x-4-\left(\frac{60x^{2}}{\frac{1}{2}}+\frac{-18x}{\frac{1}{2}}+\frac{-24}{\frac{1}{2}}\right)=0$$

Divide 60x^{2} by \frac{1}{2} to get 120x^{2}.

$$5x-4-\left(120x^{2}+\frac{-18x}{\frac{1}{2}}+\frac{-24}{\frac{1}{2}}\right)=0$$

Divide -18x by \frac{1}{2} to get -36x.

$$5x-4-\left(120x^{2}-36x+\frac{-24}{\frac{1}{2}}\right)=0$$

Divide -24 by \frac{1}{2} by multiplying -24 by the reciprocal of \frac{1}{2}.

$$5x-4-\left(120x^{2}-36x-24\times 2\right)=0$$

Multiply -24 and 2 to get -48.

$$5x-4-\left(120x^{2}-36x-48\right)=0$$

To find the opposite of 120x^{2}-36x-48, find the opposite of each term.

$$5x-4-120x^{2}+36x+48=0$$

Combine 5x and 36x to get 41x.

$$41x-4-120x^{2}+48=0$$

Add -4 and 48 to get 44.

$$41x+44-120x^{2}=0$$

Subtract 44 from both sides. Anything subtracted from zero gives its negation.

$$41x-120x^{2}=-44$$

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.

$$-120x^{2}+41x=-44$$

Divide both sides by -120.

$$\frac{-120x^{2}+41x}{-120}=-\frac{44}{-120}$$

Dividing by -120 undoes the multiplication by -120.

$$x^{2}+\frac{41}{-120}x=-\frac{44}{-120}$$

Divide 41 by -120.

$$x^{2}-\frac{41}{120}x=-\frac{44}{-120}$$

Reduce the fraction \frac{-44}{-120} to lowest terms by extracting and canceling out 4.

$$x^{2}-\frac{41}{120}x=\frac{11}{30}$$

Divide -\frac{41}{120}, the coefficient of the x term, by 2 to get -\frac{41}{240}. Then add the square of -\frac{41}{240} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{41}{120}x+\left(-\frac{41}{240}\right)^{2}=\frac{11}{30}+\left(-\frac{41}{240}\right)^{2}$$

Square -\frac{41}{240} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{41}{120}x+\frac{1681}{57600}=\frac{11}{30}+\frac{1681}{57600}$$

Add \frac{11}{30} to \frac{1681}{57600} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{41}{120}x+\frac{1681}{57600}=\frac{22801}{57600}$$

Factor x^{2}-\frac{41}{120}x+\frac{1681}{57600}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{41}{240}\right)^{2}=\frac{22801}{57600}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{41}{240}\right)^{2}}=\sqrt{\frac{22801}{57600}}$$

Simplify.

$$x-\frac{41}{240}=\frac{151}{240} x-\frac{41}{240}=-\frac{151}{240}$$

Add \frac{41}{240} to both sides of the equation.

$$x=\frac{4}{5} x=-\frac{11}{24}$$