Solve left(2x-3right)^2-5(2x-3)+6=0

$${ \left(2x-3 \right) }^{ 2 } -5(2x-3)+6=0$$

Solve for x

$$x = \frac{5}{2} = 2\frac{1}{2} = 2.5$$

$$x=3$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-3\right)^{2}.

$$4x^{2}-12x+9-5\left(2x-3\right)+6=0$$

Use the distributive property to multiply -5 by 2x-3.

$$4x^{2}-12x+9-10x+15+6=0$$

Combine -12x and -10x to get -22x.

$$4x^{2}-22x+9+15+6=0$$

Add 9 and 15 to get 24.

$$4x^{2}-22x+24+6=0$$

Add 24 and 6 to get 30.

$$4x^{2}-22x+30=0$$

Divide both sides by 2.

$$2x^{2}-11x+15=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2x^{2}+ax+bx+15. To find a and b, set up a system to be solved.

$$a+b=-11 ab=2\times 15=30$$

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 30.

$$-1,-30 -2,-15 -3,-10 -5,-6$$

Calculate the sum for each pair.

$$-1-30=-31 -2-15=-17 -3-10=-13 -5-6=-11$$

The solution is the pair that gives sum -11.

$$a=-6 b=-5$$

Rewrite 2x^{2}-11x+15 as \left(2x^{2}-6x\right)+\left(-5x+15\right).

$$\left(2x^{2}-6x\right)+\left(-5x+15\right)$$

Factor out 2x in the first and -5 in the second group.

$$2x\left(x-3\right)-5\left(x-3\right)$$

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

$$\left(x-3\right)\left(2x-5\right)$$

To find equation solutions, solve x-3=0 and 2x-5=0.

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

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-3\right)^{2}.

$$4x^{2}-12x+9-5\left(2x-3\right)+6=0$$

Use the distributive property to multiply -5 by 2x-3.

$$4x^{2}-12x+9-10x+15+6=0$$

Combine -12x and -10x to get -22x.

$$4x^{2}-22x+9+15+6=0$$

Add 9 and 15 to get 24.

$$4x^{2}-22x+24+6=0$$

Add 24 and 6 to get 30.

$$4x^{2}-22x+30=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -22 for b, and 30 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-22\right)±\sqrt{\left(-22\right)^{2}-4\times 4\times 30}}{2\times 4}$$

Square -22.

$$x=\frac{-\left(-22\right)±\sqrt{484-4\times 4\times 30}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-22\right)±\sqrt{484-16\times 30}}{2\times 4}$$

Multiply -16 times 30.

$$x=\frac{-\left(-22\right)±\sqrt{484-480}}{2\times 4}$$

Add 484 to -480.

$$x=\frac{-\left(-22\right)±\sqrt{4}}{2\times 4}$$

Take the square root of 4.

$$x=\frac{-\left(-22\right)±2}{2\times 4}$$

The opposite of -22 is 22.

$$x=\frac{22±2}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{22±2}{8}$$

Now solve the equation x=\frac{22±2}{8} when ± is plus. Add 22 to 2.

$$x=\frac{24}{8}$$

Divide 24 by 8.

$$x=3$$

Now solve the equation x=\frac{22±2}{8} when ± is minus. Subtract 2 from 22.

$$x=\frac{20}{8}$$

Reduce the fraction \frac{20}{8} to lowest terms by extracting and canceling out 4.

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

The equation is now solved.

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

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-3\right)^{2}.

$$4x^{2}-12x+9-5\left(2x-3\right)+6=0$$

Use the distributive property to multiply -5 by 2x-3.

$$4x^{2}-12x+9-10x+15+6=0$$

Combine -12x and -10x to get -22x.

$$4x^{2}-22x+9+15+6=0$$

Add 9 and 15 to get 24.

$$4x^{2}-22x+24+6=0$$

Add 24 and 6 to get 30.

$$4x^{2}-22x+30=0$$

Subtract 30 from both sides. Anything subtracted from zero gives its negation.

$$4x^{2}-22x=-30$$

Divide both sides by 4.

$$\frac{4x^{2}-22x}{4}=-\frac{30}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\left(-\frac{22}{4}\right)x=-\frac{30}{4}$$

Reduce the fraction \frac{-22}{4} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{11}{2}x=-\frac{30}{4}$$

Reduce the fraction \frac{-30}{4} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{11}{2}x=-\frac{15}{2}$$

Divide -\frac{11}{2}, the coefficient of the x term, by 2 to get -\frac{11}{4}. Then add the square of -\frac{11}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{11}{2}x+\left(-\frac{11}{4}\right)^{2}=-\frac{15}{2}+\left(-\frac{11}{4}\right)^{2}$$

Square -\frac{11}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{11}{2}x+\frac{121}{16}=-\frac{15}{2}+\frac{121}{16}$$

Add -\frac{15}{2} to \frac{121}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{11}{2}x+\frac{121}{16}=\frac{1}{16}$$

Factor x^{2}-\frac{11}{2}x+\frac{121}{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}.

$$\left(x-\frac{11}{4}\right)^{2}=\frac{1}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{11}{4}\right)^{2}}=\sqrt{\frac{1}{16}}$$

Simplify.

$$x-\frac{11}{4}=\frac{1}{4} x-\frac{11}{4}=-\frac{1}{4}$$

Add \frac{11}{4} to both sides of the equation.

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