Solve x-2x+3/3=3x^2-7x/6

$$x - \frac { 2 x + 3 } { 3 } = \frac { 3 x ^ { 2 } - 7 x } { 6 }$$

Solve for x

$$x=1$$

$$x=2$$

Steps Using Factoring By Grouping

Multiply both sides of the equation by 6, the least common multiple of 3,6.

$$6x-2\left(2x+3\right)=3x^{2}-7x$$

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

$$6x-4x-6=3x^{2}-7x$$

Combine 6x and -4x to get 2x.

$$2x-6=3x^{2}-7x$$

Subtract 3x^{2} from both sides.

$$2x-6-3x^{2}=-7x$$

Add 7x to both sides.

$$2x-6-3x^{2}+7x=0$$

Combine 2x and 7x to get 9x.

$$9x-6-3x^{2}=0$$

Divide both sides by 3.

$$3x-2-x^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

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

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

$$a+b=3 ab=-\left(-2\right)=2$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. The only such pair is the system solution.

$$a=2 b=1$$

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

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

Factor out -x in -x^{2}+2x.

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

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

$$\left(x-2\right)\left(-x+1\right)$$

To find equation solutions, solve x-2=0 and -x+1=0.

$$x=2 x=1$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 6, the least common multiple of 3,6.

$$6x-2\left(2x+3\right)=3x^{2}-7x$$

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

$$6x-4x-6=3x^{2}-7x$$

Combine 6x and -4x to get 2x.

$$2x-6=3x^{2}-7x$$

Subtract 3x^{2} from both sides.

$$2x-6-3x^{2}=-7x$$

Add 7x to both sides.

$$2x-6-3x^{2}+7x=0$$

Combine 2x and 7x to get 9x.

$$9x-6-3x^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-3x^{2}+9x-6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, 9 for b, and -6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-9±\sqrt{9^{2}-4\left(-3\right)\left(-6\right)}}{2\left(-3\right)}$$

Square 9.

$$x=\frac{-9±\sqrt{81-4\left(-3\right)\left(-6\right)}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-9±\sqrt{81+12\left(-6\right)}}{2\left(-3\right)}$$

Multiply 12 times -6.

$$x=\frac{-9±\sqrt{81-72}}{2\left(-3\right)}$$

Add 81 to -72.

$$x=\frac{-9±\sqrt{9}}{2\left(-3\right)}$$

Take the square root of 9.

$$x=\frac{-9±3}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{-9±3}{-6}$$

Now solve the equation x=\frac{-9±3}{-6} when ± is plus. Add -9 to 3.

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

Divide -6 by -6.

$$x=1$$

Now solve the equation x=\frac{-9±3}{-6} when ± is minus. Subtract 3 from -9.

$$x=-\frac{12}{-6}$$

Divide -12 by -6.

$$x=2$$

The equation is now solved.

$$x=1 x=2$$

Steps for Completing the Square

Multiply both sides of the equation by 6, the least common multiple of 3,6.

$$6x-2\left(2x+3\right)=3x^{2}-7x$$

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

$$6x-4x-6=3x^{2}-7x$$

Combine 6x and -4x to get 2x.

$$2x-6=3x^{2}-7x$$

Subtract 3x^{2} from both sides.

$$2x-6-3x^{2}=-7x$$

Add 7x to both sides.

$$2x-6-3x^{2}+7x=0$$

Combine 2x and 7x to get 9x.

$$9x-6-3x^{2}=0$$

Add 6 to both sides. Anything plus zero gives itself.

$$9x-3x^{2}=6$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-3x^{2}+9x=6$$

Divide both sides by -3.

$$\frac{-3x^{2}+9x}{-3}=\frac{6}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$x^{2}+\frac{9}{-3}x=\frac{6}{-3}$$

Divide 9 by -3.

$$x^{2}-3x=\frac{6}{-3}$$

Divide 6 by -3.

$$x^{2}-3x=-2$$

Divide -3, the coefficient of the x term, by 2 to get -\frac{3}{2}. Then add the square of -\frac{3}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-3x+\frac{9}{4}=-2+\frac{9}{4}$$

Add -2 to \frac{9}{4}.

$$x^{2}-3x+\frac{9}{4}=\frac{1}{4}$$

Factor x^{2}-3x+\frac{9}{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}.

$$\left(x-\frac{3}{2}\right)^{2}=\frac{1}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{2}\right)^{2}}=\sqrt{\frac{1}{4}}$$

Simplify.

$$x-\frac{3}{2}=\frac{1}{2} x-\frac{3}{2}=-\frac{1}{2}$$

Add \frac{3}{2} to both sides of the equation.

$$x=2 x=1$$