Solve 2(x-3)(4x-1)=(4x-6)^2-4(6+x)

$$2 ( x - 3 ) ( 4 x - 1 ) = ( 4 x - 6 ) ^ { 2 } - 4 ( 6 + x )$$

Solve for x

$$x=\frac{1}{4}=0.25$$

$$x=3$$

Steps Using Factoring By Grouping

Use the distributive property to multiply 2 by x-3.

$$\left(2x-6\right)\left(4x-1\right)=\left(4x-6\right)^{2}-4\left(6+x\right)$$

Use the distributive property to multiply 2x-6 by 4x-1 and combine like terms.

$$8x^{2}-26x+6=\left(4x-6\right)^{2}-4\left(6+x\right)$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4x-6\right)^{2}.

$$8x^{2}-26x+6=16x^{2}-48x+36-4\left(6+x\right)$$

Use the distributive property to multiply -4 by 6+x.

$$8x^{2}-26x+6=16x^{2}-48x+36-24-4x$$

Subtract 24 from 36 to get 12.

$$8x^{2}-26x+6=16x^{2}-48x+12-4x$$

Combine -48x and -4x to get -52x.

$$8x^{2}-26x+6=16x^{2}-52x+12$$

Subtract 16x^{2} from both sides.

$$8x^{2}-26x+6-16x^{2}=-52x+12$$

Combine 8x^{2} and -16x^{2} to get -8x^{2}.

$$-8x^{2}-26x+6=-52x+12$$

Add 52x to both sides.

$$-8x^{2}-26x+6+52x=12$$

Combine -26x and 52x to get 26x.

$$-8x^{2}+26x+6=12$$

Subtract 12 from both sides.

$$-8x^{2}+26x+6-12=0$$

Subtract 12 from 6 to get -6.

$$-8x^{2}+26x-6=0$$

Divide both sides by 2.

$$-4x^{2}+13x-3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -4x^{2}+ax+bx-3. To find a and b, set up a system to be solved.

$$a+b=13 ab=-4\left(-3\right)=12$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 12.

$$1,12 2,6 3,4$$

Calculate the sum for each pair.

$$1+12=13 2+6=8 3+4=7$$

The solution is the pair that gives sum 13.

$$a=12 b=1$$

Rewrite -4x^{2}+13x-3 as \left(-4x^{2}+12x\right)+\left(x-3\right).

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

Factor out 4x in the first and -1 in the second group.

$$4x\left(-x+3\right)-\left(-x+3\right)$$

Factor out common term -x+3 by using distributive property.

$$\left(-x+3\right)\left(4x-1\right)$$

To find equation solutions, solve -x+3=0 and 4x-1=0.

$$x=3 x=\frac{1}{4}$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 2 by x-3.

$$\left(2x-6\right)\left(4x-1\right)=\left(4x-6\right)^{2}-4\left(6+x\right)$$

Use the distributive property to multiply 2x-6 by 4x-1 and combine like terms.

$$8x^{2}-26x+6=\left(4x-6\right)^{2}-4\left(6+x\right)$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4x-6\right)^{2}.

$$8x^{2}-26x+6=16x^{2}-48x+36-4\left(6+x\right)$$

Use the distributive property to multiply -4 by 6+x.

$$8x^{2}-26x+6=16x^{2}-48x+36-24-4x$$

Subtract 24 from 36 to get 12.

$$8x^{2}-26x+6=16x^{2}-48x+12-4x$$

Combine -48x and -4x to get -52x.

$$8x^{2}-26x+6=16x^{2}-52x+12$$

Subtract 16x^{2} from both sides.

$$8x^{2}-26x+6-16x^{2}=-52x+12$$

Combine 8x^{2} and -16x^{2} to get -8x^{2}.

$$-8x^{2}-26x+6=-52x+12$$

Add 52x to both sides.

$$-8x^{2}-26x+6+52x=12$$

Combine -26x and 52x to get 26x.

$$-8x^{2}+26x+6=12$$

Subtract 12 from both sides.

$$-8x^{2}+26x+6-12=0$$

Subtract 12 from 6 to get -6.

$$-8x^{2}+26x-6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -8 for a, 26 for b, and -6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-26±\sqrt{26^{2}-4\left(-8\right)\left(-6\right)}}{2\left(-8\right)}$$

Square 26.

$$x=\frac{-26±\sqrt{676-4\left(-8\right)\left(-6\right)}}{2\left(-8\right)}$$

Multiply -4 times -8.

$$x=\frac{-26±\sqrt{676+32\left(-6\right)}}{2\left(-8\right)}$$

Multiply 32 times -6.

$$x=\frac{-26±\sqrt{676-192}}{2\left(-8\right)}$$

Add 676 to -192.

$$x=\frac{-26±\sqrt{484}}{2\left(-8\right)}$$

Take the square root of 484.

$$x=\frac{-26±22}{2\left(-8\right)}$$

Multiply 2 times -8.

$$x=\frac{-26±22}{-16}$$

Now solve the equation x=\frac{-26±22}{-16} when ± is plus. Add -26 to 22.

$$x=-\frac{4}{-16}$$

Reduce the fraction \frac{-4}{-16} to lowest terms by extracting and canceling out 4.

$$x=\frac{1}{4}$$

Now solve the equation x=\frac{-26±22}{-16} when ± is minus. Subtract 22 from -26.

$$x=-\frac{48}{-16}$$

Divide -48 by -16.

$$x=3$$

The equation is now solved.

$$x=\frac{1}{4} x=3$$

Steps for Completing the Square

Use the distributive property to multiply 2 by x-3.

$$\left(2x-6\right)\left(4x-1\right)=\left(4x-6\right)^{2}-4\left(6+x\right)$$

Use the distributive property to multiply 2x-6 by 4x-1 and combine like terms.

$$8x^{2}-26x+6=\left(4x-6\right)^{2}-4\left(6+x\right)$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4x-6\right)^{2}.

$$8x^{2}-26x+6=16x^{2}-48x+36-4\left(6+x\right)$$

Use the distributive property to multiply -4 by 6+x.

$$8x^{2}-26x+6=16x^{2}-48x+36-24-4x$$

Subtract 24 from 36 to get 12.

$$8x^{2}-26x+6=16x^{2}-48x+12-4x$$

Combine -48x and -4x to get -52x.

$$8x^{2}-26x+6=16x^{2}-52x+12$$

Subtract 16x^{2} from both sides.

$$8x^{2}-26x+6-16x^{2}=-52x+12$$

Combine 8x^{2} and -16x^{2} to get -8x^{2}.

$$-8x^{2}-26x+6=-52x+12$$

Add 52x to both sides.

$$-8x^{2}-26x+6+52x=12$$

Combine -26x and 52x to get 26x.

$$-8x^{2}+26x+6=12$$

Subtract 6 from both sides.

$$-8x^{2}+26x=12-6$$

Subtract 6 from 12 to get 6.

$$-8x^{2}+26x=6$$

Divide both sides by -8.

$$\frac{-8x^{2}+26x}{-8}=\frac{6}{-8}$$

Dividing by -8 undoes the multiplication by -8.

$$x^{2}+\frac{26}{-8}x=\frac{6}{-8}$$

Reduce the fraction \frac{26}{-8} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{13}{4}x=\frac{6}{-8}$$

Reduce the fraction \frac{6}{-8} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{13}{4}x=-\frac{3}{4}$$

Divide -\frac{13}{4}, the coefficient of the x term, by 2 to get -\frac{13}{8}. Then add the square of -\frac{13}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{13}{4}x+\left(-\frac{13}{8}\right)^{2}=-\frac{3}{4}+\left(-\frac{13}{8}\right)^{2}$$

Square -\frac{13}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{13}{4}x+\frac{169}{64}=-\frac{3}{4}+\frac{169}{64}$$

Add -\frac{3}{4} to \frac{169}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{13}{4}x+\frac{169}{64}=\frac{121}{64}$$

Factor x^{2}-\frac{13}{4}x+\frac{169}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{13}{8}\right)^{2}=\frac{121}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{13}{8}\right)^{2}}=\sqrt{\frac{121}{64}}$$

Simplify.

$$x-\frac{13}{8}=\frac{11}{8} x-\frac{13}{8}=-\frac{11}{8}$$

Add \frac{13}{8} to both sides of the equation.

$$x=3 x=\frac{1}{4}$$