Solve (2x-3)(-4+x)=2x(-4+x)-3x(-4+x)

$$( 2 x - 3 ) ( - 4 + x ) = 2 x ( - 4 + x ) - 3 x ( - 4 + x )$$

Solve for x

$$x=1$$

$$x=4$$

Steps Using Factoring By Grouping

Use the distributive property to multiply 2x-3 by -4+x and combine like terms.

$$-11x+2x^{2}+12=2x\left(-4+x\right)-3x\left(-4+x\right)$$

Use the distributive property to multiply 2x by -4+x.

$$-11x+2x^{2}+12=-8x+2x^{2}-3x\left(-4+x\right)$$

Add 3x\left(-4+x\right) to both sides.

$$-11x+2x^{2}+12+3x\left(-4+x\right)=-8x+2x^{2}$$

Use the distributive property to multiply 3x by -4+x.

$$-11x+2x^{2}+12-12x+3x^{2}=-8x+2x^{2}$$

Combine -11x and -12x to get -23x.

$$-23x+2x^{2}+12+3x^{2}=-8x+2x^{2}$$

Combine 2x^{2} and 3x^{2} to get 5x^{2}.

$$-23x+5x^{2}+12=-8x+2x^{2}$$

Add 8x to both sides.

$$-23x+5x^{2}+12+8x=2x^{2}$$

Combine -23x and 8x to get -15x.

$$-15x+5x^{2}+12=2x^{2}$$

Subtract 2x^{2} from both sides.

$$-15x+5x^{2}+12-2x^{2}=0$$

Combine 5x^{2} and -2x^{2} to get 3x^{2}.

$$-15x+3x^{2}+12=0$$

Divide both sides by 3.

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

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

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+4. To find a and b, set up a system to be solved.

$$a+b=-5 ab=1\times 4=4$$

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 4.

$$-1,-4 -2,-2$$

Calculate the sum for each pair.

$$-1-4=-5 -2-2=-4$$

The solution is the pair that gives sum -5.

$$a=-4 b=-1$$

Rewrite x^{2}-5x+4 as \left(x^{2}-4x\right)+\left(-x+4\right).

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

Factor out x in the first and -1 in the second group.

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

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

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

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

$$x=4 x=1$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 2x-3 by -4+x and combine like terms.

$$-11x+2x^{2}+12=2x\left(-4+x\right)-3x\left(-4+x\right)$$

Use the distributive property to multiply 2x by -4+x.

$$-11x+2x^{2}+12=-8x+2x^{2}-3x\left(-4+x\right)$$

Add 3x\left(-4+x\right) to both sides.

$$-11x+2x^{2}+12+3x\left(-4+x\right)=-8x+2x^{2}$$

Use the distributive property to multiply 3x by -4+x.

$$-11x+2x^{2}+12-12x+3x^{2}=-8x+2x^{2}$$

Combine -11x and -12x to get -23x.

$$-23x+2x^{2}+12+3x^{2}=-8x+2x^{2}$$

Combine 2x^{2} and 3x^{2} to get 5x^{2}.

$$-23x+5x^{2}+12=-8x+2x^{2}$$

Add 8x to both sides.

$$-23x+5x^{2}+12+8x=2x^{2}$$

Combine -23x and 8x to get -15x.

$$-15x+5x^{2}+12=2x^{2}$$

Subtract 2x^{2} from both sides.

$$-15x+5x^{2}+12-2x^{2}=0$$

Combine 5x^{2} and -2x^{2} to get 3x^{2}.

$$-15x+3x^{2}+12=0$$

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.

$$3x^{2}-15x+12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -15 for b, and 12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-15\right)±\sqrt{\left(-15\right)^{2}-4\times 3\times 12}}{2\times 3}$$

Square -15.

$$x=\frac{-\left(-15\right)±\sqrt{225-4\times 3\times 12}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-15\right)±\sqrt{225-12\times 12}}{2\times 3}$$

Multiply -12 times 12.

$$x=\frac{-\left(-15\right)±\sqrt{225-144}}{2\times 3}$$

Add 225 to -144.

$$x=\frac{-\left(-15\right)±\sqrt{81}}{2\times 3}$$

Take the square root of 81.

$$x=\frac{-\left(-15\right)±9}{2\times 3}$$

The opposite of -15 is 15.

$$x=\frac{15±9}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{15±9}{6}$$

Now solve the equation x=\frac{15±9}{6} when ± is plus. Add 15 to 9.

$$x=\frac{24}{6}$$

Divide 24 by 6.

$$x=4$$

Now solve the equation x=\frac{15±9}{6} when ± is minus. Subtract 9 from 15.

$$x=\frac{6}{6}$$

Divide 6 by 6.

$$x=1$$

The equation is now solved.

$$x=4 x=1$$

Steps for Completing the Square

Use the distributive property to multiply 2x-3 by -4+x and combine like terms.

$$-11x+2x^{2}+12=2x\left(-4+x\right)-3x\left(-4+x\right)$$

Use the distributive property to multiply 2x by -4+x.

$$-11x+2x^{2}+12=-8x+2x^{2}-3x\left(-4+x\right)$$

Add 3x\left(-4+x\right) to both sides.

$$-11x+2x^{2}+12+3x\left(-4+x\right)=-8x+2x^{2}$$

Use the distributive property to multiply 3x by -4+x.

$$-11x+2x^{2}+12-12x+3x^{2}=-8x+2x^{2}$$

Combine -11x and -12x to get -23x.

$$-23x+2x^{2}+12+3x^{2}=-8x+2x^{2}$$

Combine 2x^{2} and 3x^{2} to get 5x^{2}.

$$-23x+5x^{2}+12=-8x+2x^{2}$$

Add 8x to both sides.

$$-23x+5x^{2}+12+8x=2x^{2}$$

Combine -23x and 8x to get -15x.

$$-15x+5x^{2}+12=2x^{2}$$

Subtract 2x^{2} from both sides.

$$-15x+5x^{2}+12-2x^{2}=0$$

Combine 5x^{2} and -2x^{2} to get 3x^{2}.

$$-15x+3x^{2}+12=0$$

Subtract 12 from both sides. Anything subtracted from zero gives its negation.

$$-15x+3x^{2}=-12$$

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.

$$3x^{2}-15x=-12$$

Divide both sides by 3.

$$\frac{3x^{2}-15x}{3}=-\frac{12}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\left(-\frac{15}{3}\right)x=-\frac{12}{3}$$

Divide -15 by 3.

$$x^{2}-5x=-\frac{12}{3}$$

Divide -12 by 3.

$$x^{2}-5x=-4$$

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

$$x^{2}-5x+\left(-\frac{5}{2}\right)^{2}=-4+\left(-\frac{5}{2}\right)^{2}$$

Square -\frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-5x+\frac{25}{4}=-4+\frac{25}{4}$$

Add -4 to \frac{25}{4}.

$$x^{2}-5x+\frac{25}{4}=\frac{9}{4}$$

Factor x^{2}-5x+\frac{25}{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}.

$$\left(x-\frac{5}{2}\right)^{2}=\frac{9}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{2}\right)^{2}}=\sqrt{\frac{9}{4}}$$

Simplify.

$$x-\frac{5}{2}=\frac{3}{2} x-\frac{5}{2}=-\frac{3}{2}$$

Add \frac{5}{2} to both sides of the equation.

$$x=4 x=1$$