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

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

Solve for x

$$x=2$$

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

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}.

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

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

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

Combine -12x and -6x to get -18x.

$$4x^{2}-18x+9+9+2=0$$

Add 9 and 9 to get 18.

$$4x^{2}-18x+18+2=0$$

Add 18 and 2 to get 20.

$$4x^{2}-18x+20=0$$

Divide both sides by 2.

$$2x^{2}-9x+10=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2x^{2}+ax+bx+10. To find a and b, set up a system to be solved.

$$a+b=-9 ab=2\times 10=20$$

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 20.

$$-1,-20 -2,-10 -4,-5$$

Calculate the sum for each pair.

$$-1-20=-21 -2-10=-12 -4-5=-9$$

The solution is the pair that gives sum -9.

$$a=-5 b=-4$$

Rewrite 2x^{2}-9x+10 as \left(2x^{2}-5x\right)+\left(-4x+10\right).

$$\left(2x^{2}-5x\right)+\left(-4x+10\right)$$

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

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

Factor out common term 2x-5 by using distributive property.

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

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

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

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}.

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

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

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

Combine -12x and -6x to get -18x.

$$4x^{2}-18x+9+9+2=0$$

Add 9 and 9 to get 18.

$$4x^{2}-18x+18+2=0$$

Add 18 and 2 to get 20.

$$4x^{2}-18x+20=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -18 for b, and 20 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-18\right)±\sqrt{\left(-18\right)^{2}-4\times 4\times 20}}{2\times 4}$$

Square -18.

$$x=\frac{-\left(-18\right)±\sqrt{324-4\times 4\times 20}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-18\right)±\sqrt{324-16\times 20}}{2\times 4}$$

Multiply -16 times 20.

$$x=\frac{-\left(-18\right)±\sqrt{324-320}}{2\times 4}$$

Add 324 to -320.

$$x=\frac{-\left(-18\right)±\sqrt{4}}{2\times 4}$$

Take the square root of 4.

$$x=\frac{-\left(-18\right)±2}{2\times 4}$$

The opposite of -18 is 18.

$$x=\frac{18±2}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{18±2}{8}$$

Now solve the equation x=\frac{18±2}{8} when ± is plus. Add 18 to 2.

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

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

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

Now solve the equation x=\frac{18±2}{8} when ± is minus. Subtract 2 from 18.

$$x=\frac{16}{8}$$

Divide 16 by 8.

$$x=2$$

The equation is now solved.

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

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}.

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

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

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

Combine -12x and -6x to get -18x.

$$4x^{2}-18x+9+9+2=0$$

Add 9 and 9 to get 18.

$$4x^{2}-18x+18+2=0$$

Add 18 and 2 to get 20.

$$4x^{2}-18x+20=0$$

Subtract 20 from both sides. Anything subtracted from zero gives its negation.

$$4x^{2}-18x=-20$$

Divide both sides by 4.

$$\frac{4x^{2}-18x}{4}=-\frac{20}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\left(-\frac{18}{4}\right)x=-\frac{20}{4}$$

Reduce the fraction \frac{-18}{4} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{9}{2}x=-\frac{20}{4}$$

Divide -20 by 4.

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

Divide -\frac{9}{2}, the coefficient of the x term, by 2 to get -\frac{9}{4}. Then add the square of -\frac{9}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{9}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{9}{2}x+\frac{81}{16}=-5+\frac{81}{16}$$

Add -5 to \frac{81}{16}.

$$x^{2}-\frac{9}{2}x+\frac{81}{16}=\frac{1}{16}$$

Factor x^{2}-\frac{9}{2}x+\frac{81}{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}.

$$\left(x-\frac{9}{4}\right)^{2}=\frac{1}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{9}{4}\right)^{2}}=\sqrt{\frac{1}{16}}$$

Simplify.

$$x-\frac{9}{4}=\frac{1}{4} x-\frac{9}{4}=-\frac{1}{4}$$

Add \frac{9}{4} to both sides of the equation.

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