Solve (4x-7)(2x-4)=84+4x^2-32x

$$( 4 x - 7 ) ( 2 x - 4 ) = 84 + 4 x ^ { 2 } - 32 x$$

Solve for x

$$x=-4$$

$$x = \frac{7}{2} = 3\frac{1}{2} = 3.5$$

Steps Using Factoring By Grouping

Use the distributive property to multiply 4x-7 by 2x-4 and combine like terms.

$$8x^{2}-30x+28=84+4x^{2}-32x$$

Subtract 84 from both sides.

$$8x^{2}-30x+28-84=4x^{2}-32x$$

Subtract 84 from 28 to get -56.

$$8x^{2}-30x-56=4x^{2}-32x$$

Subtract 4x^{2} from both sides.

$$8x^{2}-30x-56-4x^{2}=-32x$$

Combine 8x^{2} and -4x^{2} to get 4x^{2}.

$$4x^{2}-30x-56=-32x$$

Add 32x to both sides.

$$4x^{2}-30x-56+32x=0$$

Combine -30x and 32x to get 2x.

$$4x^{2}+2x-56=0$$

Divide both sides by 2.

$$2x^{2}+x-28=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2x^{2}+ax+bx-28. To find a and b, set up a system to be solved.

$$a+b=1 ab=2\left(-28\right)=-56$$

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 -56.

$$-1,56 -2,28 -4,14 -7,8$$

Calculate the sum for each pair.

$$-1+56=55 -2+28=26 -4+14=10 -7+8=1$$

The solution is the pair that gives sum 1.

$$a=-7 b=8$$

Rewrite 2x^{2}+x-28 as \left(2x^{2}-7x\right)+\left(8x-28\right).

$$\left(2x^{2}-7x\right)+\left(8x-28\right)$$

Factor out x in the first and 4 in the second group.

$$x\left(2x-7\right)+4\left(2x-7\right)$$

Factor out common term 2x-7 by using distributive property.

$$\left(2x-7\right)\left(x+4\right)$$

To find equation solutions, solve 2x-7=0 and x+4=0.

$$x=\frac{7}{2} x=-4$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 4x-7 by 2x-4 and combine like terms.

$$8x^{2}-30x+28=84+4x^{2}-32x$$

Subtract 84 from both sides.

$$8x^{2}-30x+28-84=4x^{2}-32x$$

Subtract 84 from 28 to get -56.

$$8x^{2}-30x-56=4x^{2}-32x$$

Subtract 4x^{2} from both sides.

$$8x^{2}-30x-56-4x^{2}=-32x$$

Combine 8x^{2} and -4x^{2} to get 4x^{2}.

$$4x^{2}-30x-56=-32x$$

Add 32x to both sides.

$$4x^{2}-30x-56+32x=0$$

Combine -30x and 32x to get 2x.

$$4x^{2}+2x-56=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, 2 for b, and -56 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-2±\sqrt{2^{2}-4\times 4\left(-56\right)}}{2\times 4}$$

Square 2.

$$x=\frac{-2±\sqrt{4-4\times 4\left(-56\right)}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-2±\sqrt{4-16\left(-56\right)}}{2\times 4}$$

Multiply -16 times -56.

$$x=\frac{-2±\sqrt{4+896}}{2\times 4}$$

Add 4 to 896.

$$x=\frac{-2±\sqrt{900}}{2\times 4}$$

Take the square root of 900.

$$x=\frac{-2±30}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-2±30}{8}$$

Now solve the equation x=\frac{-2±30}{8} when ± is plus. Add -2 to 30.

$$x=\frac{28}{8}$$

Reduce the fraction \frac{28}{8} to lowest terms by extracting and canceling out 4.

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

Now solve the equation x=\frac{-2±30}{8} when ± is minus. Subtract 30 from -2.

$$x=-\frac{32}{8}$$

Divide -32 by 8.

$$x=-4$$

The equation is now solved.

$$x=\frac{7}{2} x=-4$$

Steps for Completing the Square

Use the distributive property to multiply 4x-7 by 2x-4 and combine like terms.

$$8x^{2}-30x+28=84+4x^{2}-32x$$

Subtract 4x^{2} from both sides.

$$8x^{2}-30x+28-4x^{2}=84-32x$$

Combine 8x^{2} and -4x^{2} to get 4x^{2}.

$$4x^{2}-30x+28=84-32x$$

Add 32x to both sides.

$$4x^{2}-30x+28+32x=84$$

Combine -30x and 32x to get 2x.

$$4x^{2}+2x+28=84$$

Subtract 28 from both sides.

$$4x^{2}+2x=84-28$$

Subtract 28 from 84 to get 56.

$$4x^{2}+2x=56$$

Divide both sides by 4.

$$\frac{4x^{2}+2x}{4}=\frac{56}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\frac{2}{4}x=\frac{56}{4}$$

Reduce the fraction \frac{2}{4} to lowest terms by extracting and canceling out 2.

$$x^{2}+\frac{1}{2}x=\frac{56}{4}$$

Divide 56 by 4.

$$x^{2}+\frac{1}{2}x=14$$

Divide \frac{1}{2}, the coefficient of the x term, by 2 to get \frac{1}{4}. Then add the square of \frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{1}{2}x+\left(\frac{1}{4}\right)^{2}=14+\left(\frac{1}{4}\right)^{2}$$

Square \frac{1}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=14+\frac{1}{16}$$

Add 14 to \frac{1}{16}.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{225}{16}$$

Factor x^{2}+\frac{1}{2}x+\frac{1}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{4}\right)^{2}=\frac{225}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{4}\right)^{2}}=\sqrt{\frac{225}{16}}$$

Simplify.

$$x+\frac{1}{4}=\frac{15}{4} x+\frac{1}{4}=-\frac{15}{4}$$

Subtract \frac{1}{4} from both sides of the equation.

$$x=\frac{7}{2} x=-4$$