Solve (5x-4)^2-(3x+5)(2x-1)=20x(x-2)+27

$$( 5 x - 4 ) ^ { 2 } - ( 3 x + 5 ) ( 2 x - 1 ) = 20 x ( x - 2 ) + 27$$

Solve for x

$$x=-6$$

$$x=-1$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5x-4\right)^{2}.

$$25x^{2}-40x+16-\left(3x+5\right)\left(2x-1\right)=20x\left(x-2\right)+27$$

Use the distributive property to multiply 3x+5 by 2x-1 and combine like terms.

$$25x^{2}-40x+16-\left(6x^{2}+7x-5\right)=20x\left(x-2\right)+27$$

To find the opposite of 6x^{2}+7x-5, find the opposite of each term.

$$25x^{2}-40x+16-6x^{2}-7x+5=20x\left(x-2\right)+27$$

Combine 25x^{2} and -6x^{2} to get 19x^{2}.

$$19x^{2}-40x+16-7x+5=20x\left(x-2\right)+27$$

Combine -40x and -7x to get -47x.

$$19x^{2}-47x+16+5=20x\left(x-2\right)+27$$

Add 16 and 5 to get 21.

$$19x^{2}-47x+21=20x\left(x-2\right)+27$$

Use the distributive property to multiply 20x by x-2.

$$19x^{2}-47x+21=20x^{2}-40x+27$$

Subtract 20x^{2} from both sides.

$$19x^{2}-47x+21-20x^{2}=-40x+27$$

Combine 19x^{2} and -20x^{2} to get -x^{2}.

$$-x^{2}-47x+21=-40x+27$$

Add 40x to both sides.

$$-x^{2}-47x+21+40x=27$$

Combine -47x and 40x to get -7x.

$$-x^{2}-7x+21=27$$

Subtract 27 from both sides.

$$-x^{2}-7x+21-27=0$$

Subtract 27 from 21 to get -6.

$$-x^{2}-7x-6=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-6. To find a and b, set up a system to be solved.

$$a+b=-7 ab=-\left(-6\right)=6$$

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 6.

$$-1,-6 -2,-3$$

Calculate the sum for each pair.

$$-1-6=-7 -2-3=-5$$

The solution is the pair that gives sum -7.

$$a=-1 b=-6$$

Rewrite -x^{2}-7x-6 as \left(-x^{2}-x\right)+\left(-6x-6\right).

$$\left(-x^{2}-x\right)+\left(-6x-6\right)$$

Factor out x in the first and 6 in the second group.

$$x\left(-x-1\right)+6\left(-x-1\right)$$

Factor out common term -x-1 by using distributive property.

$$\left(-x-1\right)\left(x+6\right)$$

To find equation solutions, solve -x-1=0 and x+6=0.

$$x=-1 x=-6$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5x-4\right)^{2}.

$$25x^{2}-40x+16-\left(3x+5\right)\left(2x-1\right)=20x\left(x-2\right)+27$$

Use the distributive property to multiply 3x+5 by 2x-1 and combine like terms.

$$25x^{2}-40x+16-\left(6x^{2}+7x-5\right)=20x\left(x-2\right)+27$$

To find the opposite of 6x^{2}+7x-5, find the opposite of each term.

$$25x^{2}-40x+16-6x^{2}-7x+5=20x\left(x-2\right)+27$$

Combine 25x^{2} and -6x^{2} to get 19x^{2}.

$$19x^{2}-40x+16-7x+5=20x\left(x-2\right)+27$$

Combine -40x and -7x to get -47x.

$$19x^{2}-47x+16+5=20x\left(x-2\right)+27$$

Add 16 and 5 to get 21.

$$19x^{2}-47x+21=20x\left(x-2\right)+27$$

Use the distributive property to multiply 20x by x-2.

$$19x^{2}-47x+21=20x^{2}-40x+27$$

Subtract 20x^{2} from both sides.

$$19x^{2}-47x+21-20x^{2}=-40x+27$$

Combine 19x^{2} and -20x^{2} to get -x^{2}.

$$-x^{2}-47x+21=-40x+27$$

Add 40x to both sides.

$$-x^{2}-47x+21+40x=27$$

Combine -47x and 40x to get -7x.

$$-x^{2}-7x+21=27$$

Subtract 27 from both sides.

$$-x^{2}-7x+21-27=0$$

Subtract 27 from 21 to get -6.

$$-x^{2}-7x-6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -7 for b, and -6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\left(-1\right)\left(-6\right)}}{2\left(-1\right)}$$

Square -7.

$$x=\frac{-\left(-7\right)±\sqrt{49-4\left(-1\right)\left(-6\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-7\right)±\sqrt{49+4\left(-6\right)}}{2\left(-1\right)}$$

Multiply 4 times -6.

$$x=\frac{-\left(-7\right)±\sqrt{49-24}}{2\left(-1\right)}$$

Add 49 to -24.

$$x=\frac{-\left(-7\right)±\sqrt{25}}{2\left(-1\right)}$$

Take the square root of 25.

$$x=\frac{-\left(-7\right)±5}{2\left(-1\right)}$$

The opposite of -7 is 7.

$$x=\frac{7±5}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{7±5}{-2}$$

Now solve the equation x=\frac{7±5}{-2} when ± is plus. Add 7 to 5.

$$x=\frac{12}{-2}$$

Divide 12 by -2.

$$x=-6$$

Now solve the equation x=\frac{7±5}{-2} when ± is minus. Subtract 5 from 7.

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

Divide 2 by -2.

$$x=-1$$

The equation is now solved.

$$x=-6 x=-1$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5x-4\right)^{2}.

$$25x^{2}-40x+16-\left(3x+5\right)\left(2x-1\right)=20x\left(x-2\right)+27$$

Use the distributive property to multiply 3x+5 by 2x-1 and combine like terms.

$$25x^{2}-40x+16-\left(6x^{2}+7x-5\right)=20x\left(x-2\right)+27$$

To find the opposite of 6x^{2}+7x-5, find the opposite of each term.

$$25x^{2}-40x+16-6x^{2}-7x+5=20x\left(x-2\right)+27$$

Combine 25x^{2} and -6x^{2} to get 19x^{2}.

$$19x^{2}-40x+16-7x+5=20x\left(x-2\right)+27$$

Combine -40x and -7x to get -47x.

$$19x^{2}-47x+16+5=20x\left(x-2\right)+27$$

Add 16 and 5 to get 21.

$$19x^{2}-47x+21=20x\left(x-2\right)+27$$

Use the distributive property to multiply 20x by x-2.

$$19x^{2}-47x+21=20x^{2}-40x+27$$

Subtract 20x^{2} from both sides.

$$19x^{2}-47x+21-20x^{2}=-40x+27$$

Combine 19x^{2} and -20x^{2} to get -x^{2}.

$$-x^{2}-47x+21=-40x+27$$

Add 40x to both sides.

$$-x^{2}-47x+21+40x=27$$

Combine -47x and 40x to get -7x.

$$-x^{2}-7x+21=27$$

Subtract 21 from both sides.

$$-x^{2}-7x=27-21$$

Subtract 21 from 27 to get 6.

$$-x^{2}-7x=6$$

Divide both sides by -1.

$$\frac{-x^{2}-7x}{-1}=\frac{6}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\left(-\frac{7}{-1}\right)x=\frac{6}{-1}$$

Divide -7 by -1.

$$x^{2}+7x=\frac{6}{-1}$$

Divide 6 by -1.

$$x^{2}+7x=-6$$

Divide 7, the coefficient of the x term, by 2 to get \frac{7}{2}. Then add the square of \frac{7}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+7x+\left(\frac{7}{2}\right)^{2}=-6+\left(\frac{7}{2}\right)^{2}$$

Square \frac{7}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+7x+\frac{49}{4}=-6+\frac{49}{4}$$

Add -6 to \frac{49}{4}.

$$x^{2}+7x+\frac{49}{4}=\frac{25}{4}$$

Factor x^{2}+7x+\frac{49}{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}.

$$\left(x+\frac{7}{2}\right)^{2}=\frac{25}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{7}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Simplify.

$$x+\frac{7}{2}=\frac{5}{2} x+\frac{7}{2}=-\frac{5}{2}$$

Subtract \frac{7}{2} from both sides of the equation.

$$x=-1 x=-6$$