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

$$( 2 x - 3 ) ( 3 x - 1 ) - ( x + 3 ) ( x - 2 ) = 6 x$$

Solve for x

$$x=\frac{3}{5}=0.6$$

$$x=3$$

Steps Using Factoring By Grouping

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

$$6x^{2}-11x+3-\left(x+3\right)\left(x-2\right)=6x$$

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

$$6x^{2}-11x+3-\left(x^{2}+x-6\right)=6x$$

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

$$6x^{2}-11x+3-x^{2}-x+6=6x$$

Combine 6x^{2} and -x^{2} to get 5x^{2}.

$$5x^{2}-11x+3-x+6=6x$$

Combine -11x and -x to get -12x.

$$5x^{2}-12x+3+6=6x$$

Add 3 and 6 to get 9.

$$5x^{2}-12x+9=6x$$

Subtract 6x from both sides.

$$5x^{2}-12x+9-6x=0$$

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

$$5x^{2}-18x+9=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 5x^{2}+ax+bx+9. To find a and b, set up a system to be solved.

$$a+b=-18 ab=5\times 9=45$$

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 45.

$$-1,-45 -3,-15 -5,-9$$

Calculate the sum for each pair.

$$-1-45=-46 -3-15=-18 -5-9=-14$$

The solution is the pair that gives sum -18.

$$a=-15 b=-3$$

Rewrite 5x^{2}-18x+9 as \left(5x^{2}-15x\right)+\left(-3x+9\right).

$$\left(5x^{2}-15x\right)+\left(-3x+9\right)$$

Factor out 5x in the first and -3 in the second group.

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

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

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

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

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

Steps Using the Quadratic Formula

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

$$6x^{2}-11x+3-\left(x+3\right)\left(x-2\right)=6x$$

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

$$6x^{2}-11x+3-\left(x^{2}+x-6\right)=6x$$

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

$$6x^{2}-11x+3-x^{2}-x+6=6x$$

Combine 6x^{2} and -x^{2} to get 5x^{2}.

$$5x^{2}-11x+3-x+6=6x$$

Combine -11x and -x to get -12x.

$$5x^{2}-12x+3+6=6x$$

Add 3 and 6 to get 9.

$$5x^{2}-12x+9=6x$$

Subtract 6x from both sides.

$$5x^{2}-12x+9-6x=0$$

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

$$5x^{2}-18x+9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -18 for b, and 9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-18\right)±\sqrt{\left(-18\right)^{2}-4\times 5\times 9}}{2\times 5}$$

Square -18.

$$x=\frac{-\left(-18\right)±\sqrt{324-4\times 5\times 9}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-\left(-18\right)±\sqrt{324-20\times 9}}{2\times 5}$$

Multiply -20 times 9.

$$x=\frac{-\left(-18\right)±\sqrt{324-180}}{2\times 5}$$

Add 324 to -180.

$$x=\frac{-\left(-18\right)±\sqrt{144}}{2\times 5}$$

Take the square root of 144.

$$x=\frac{-\left(-18\right)±12}{2\times 5}$$

The opposite of -18 is 18.

$$x=\frac{18±12}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{18±12}{10}$$

Now solve the equation x=\frac{18±12}{10} when ± is plus. Add 18 to 12.

$$x=\frac{30}{10}$$

Divide 30 by 10.

$$x=3$$

Now solve the equation x=\frac{18±12}{10} when ± is minus. Subtract 12 from 18.

$$x=\frac{6}{10}$$

Reduce the fraction \frac{6}{10} to lowest terms by extracting and canceling out 2.

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

The equation is now solved.

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

Steps for Completing the Square

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

$$6x^{2}-11x+3-\left(x+3\right)\left(x-2\right)=6x$$

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

$$6x^{2}-11x+3-\left(x^{2}+x-6\right)=6x$$

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

$$6x^{2}-11x+3-x^{2}-x+6=6x$$

Combine 6x^{2} and -x^{2} to get 5x^{2}.

$$5x^{2}-11x+3-x+6=6x$$

Combine -11x and -x to get -12x.

$$5x^{2}-12x+3+6=6x$$

Add 3 and 6 to get 9.

$$5x^{2}-12x+9=6x$$

Subtract 6x from both sides.

$$5x^{2}-12x+9-6x=0$$

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

$$5x^{2}-18x+9=0$$

Subtract 9 from both sides. Anything subtracted from zero gives its negation.

$$5x^{2}-18x=-9$$

Divide both sides by 5.

$$\frac{5x^{2}-18x}{5}=-\frac{9}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$x^{2}-\frac{18}{5}x=-\frac{9}{5}$$

Divide -\frac{18}{5}, the coefficient of the x term, by 2 to get -\frac{9}{5}. Then add the square of -\frac{9}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{18}{5}x+\left(-\frac{9}{5}\right)^{2}=-\frac{9}{5}+\left(-\frac{9}{5}\right)^{2}$$

Square -\frac{9}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{18}{5}x+\frac{81}{25}=-\frac{9}{5}+\frac{81}{25}$$

Add -\frac{9}{5} to \frac{81}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{18}{5}x+\frac{81}{25}=\frac{36}{25}$$

Factor x^{2}-\frac{18}{5}x+\frac{81}{25}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{9}{5}\right)^{2}=\frac{36}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{9}{5}\right)^{2}}=\sqrt{\frac{36}{25}}$$

Simplify.

$$x-\frac{9}{5}=\frac{6}{5} x-\frac{9}{5}=-\frac{6}{5}$$

Add \frac{9}{5} to both sides of the equation.

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