Solve (x-5)(3x-4)-(2x+8)(x-5)=0

$$( x - 5 ) ( 3 x - 4 ) - ( 2 x + 8 ) ( x - 5 ) = 0$$

Solve for x

$$x=12$$

$$x=5$$

Steps Using Factoring

Use the distributive property to multiply x-5 by 3x-4 and combine like terms.

$$3x^{2}-19x+20-\left(2x+8\right)\left(x-5\right)=0$$

Use the distributive property to multiply 2x+8 by x-5 and combine like terms.

$$3x^{2}-19x+20-\left(2x^{2}-2x-40\right)=0$$

To find the opposite of 2x^{2}-2x-40, find the opposite of each term.

$$3x^{2}-19x+20-2x^{2}+2x+40=0$$

Combine 3x^{2} and -2x^{2} to get x^{2}.

$$x^{2}-19x+20+2x+40=0$$

Combine -19x and 2x to get -17x.

$$x^{2}-17x+20+40=0$$

Add 20 and 40 to get 60.

$$x^{2}-17x+60=0$$

To solve the equation, factor x^{2}-17x+60 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=-17 ab=60$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 60.

$$-1,-60 -2,-30 -3,-20 -4,-15 -5,-12 -6,-10$$

Calculate the sum for each pair.

$$-1-60=-61 -2-30=-32 -3-20=-23 -4-15=-19 -5-12=-17 -6-10=-16$$

The solution is the pair that gives sum -17.

$$a=-12 b=-5$$

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

$$\left(x-12\right)\left(x-5\right)$$

To find equation solutions, solve x-12=0 and x-5=0.

$$x=12 x=5$$

Steps Using Factoring By Grouping

Use the distributive property to multiply x-5 by 3x-4 and combine like terms.

$$3x^{2}-19x+20-\left(2x+8\right)\left(x-5\right)=0$$

Use the distributive property to multiply 2x+8 by x-5 and combine like terms.

$$3x^{2}-19x+20-\left(2x^{2}-2x-40\right)=0$$

To find the opposite of 2x^{2}-2x-40, find the opposite of each term.

$$3x^{2}-19x+20-2x^{2}+2x+40=0$$

Combine 3x^{2} and -2x^{2} to get x^{2}.

$$x^{2}-19x+20+2x+40=0$$

Combine -19x and 2x to get -17x.

$$x^{2}-17x+20+40=0$$

Add 20 and 40 to get 60.

$$x^{2}-17x+60=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+60. To find a and b, set up a system to be solved.

$$a+b=-17 ab=1\times 60=60$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 60.

$$-1,-60 -2,-30 -3,-20 -4,-15 -5,-12 -6,-10$$

Calculate the sum for each pair.

$$-1-60=-61 -2-30=-32 -3-20=-23 -4-15=-19 -5-12=-17 -6-10=-16$$

The solution is the pair that gives sum -17.

$$a=-12 b=-5$$

Rewrite x^{2}-17x+60 as \left(x^{2}-12x\right)+\left(-5x+60\right).

$$\left(x^{2}-12x\right)+\left(-5x+60\right)$$

Factor out x in the first and -5 in the second group.

$$x\left(x-12\right)-5\left(x-12\right)$$

Factor out common term x-12 by using distributive property.

$$\left(x-12\right)\left(x-5\right)$$

To find equation solutions, solve x-12=0 and x-5=0.

$$x=12 x=5$$

Steps Using the Quadratic Formula

Use the distributive property to multiply x-5 by 3x-4 and combine like terms.

$$3x^{2}-19x+20-\left(2x+8\right)\left(x-5\right)=0$$

Use the distributive property to multiply 2x+8 by x-5 and combine like terms.

$$3x^{2}-19x+20-\left(2x^{2}-2x-40\right)=0$$

To find the opposite of 2x^{2}-2x-40, find the opposite of each term.

$$3x^{2}-19x+20-2x^{2}+2x+40=0$$

Combine 3x^{2} and -2x^{2} to get x^{2}.

$$x^{2}-19x+20+2x+40=0$$

Combine -19x and 2x to get -17x.

$$x^{2}-17x+20+40=0$$

Add 20 and 40 to get 60.

$$x^{2}-17x+60=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -17 for b, and 60 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-17\right)±\sqrt{\left(-17\right)^{2}-4\times 60}}{2}$$

Square -17.

$$x=\frac{-\left(-17\right)±\sqrt{289-4\times 60}}{2}$$

Multiply -4 times 60.

$$x=\frac{-\left(-17\right)±\sqrt{289-240}}{2}$$

Add 289 to -240.

$$x=\frac{-\left(-17\right)±\sqrt{49}}{2}$$

Take the square root of 49.

$$x=\frac{-\left(-17\right)±7}{2}$$

The opposite of -17 is 17.

$$x=\frac{17±7}{2}$$

Now solve the equation x=\frac{17±7}{2} when ± is plus. Add 17 to 7.

$$x=\frac{24}{2}$$

Divide 24 by 2.

$$x=12$$

Now solve the equation x=\frac{17±7}{2} when ± is minus. Subtract 7 from 17.

$$x=\frac{10}{2}$$

Divide 10 by 2.

$$x=5$$

The equation is now solved.

$$x=12 x=5$$

Steps for Completing the Square

Use the distributive property to multiply x-5 by 3x-4 and combine like terms.

$$3x^{2}-19x+20-\left(2x+8\right)\left(x-5\right)=0$$

Use the distributive property to multiply 2x+8 by x-5 and combine like terms.

$$3x^{2}-19x+20-\left(2x^{2}-2x-40\right)=0$$

To find the opposite of 2x^{2}-2x-40, find the opposite of each term.

$$3x^{2}-19x+20-2x^{2}+2x+40=0$$

Combine 3x^{2} and -2x^{2} to get x^{2}.

$$x^{2}-19x+20+2x+40=0$$

Combine -19x and 2x to get -17x.

$$x^{2}-17x+20+40=0$$

Add 20 and 40 to get 60.

$$x^{2}-17x+60=0$$

Subtract 60 from both sides. Anything subtracted from zero gives its negation.

$$x^{2}-17x=-60$$

Divide -17, the coefficient of the x term, by 2 to get -\frac{17}{2}. Then add the square of -\frac{17}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-17x+\left(-\frac{17}{2}\right)^{2}=-60+\left(-\frac{17}{2}\right)^{2}$$

Square -\frac{17}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-17x+\frac{289}{4}=-60+\frac{289}{4}$$

Add -60 to \frac{289}{4}.

$$x^{2}-17x+\frac{289}{4}=\frac{49}{4}$$

Factor x^{2}-17x+\frac{289}{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}.

$$\left(x-\frac{17}{2}\right)^{2}=\frac{49}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{17}{2}\right)^{2}}=\sqrt{\frac{49}{4}}$$

Simplify.

$$x-\frac{17}{2}=\frac{7}{2} x-\frac{17}{2}=-\frac{7}{2}$$

Add \frac{17}{2} to both sides of the equation.

$$x=12 x=5$$