Solve -6x^2+9x-6=-4x^2+x

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

Solve for x

$$x=1$$

$$x=3$$

Steps Using Factoring By Grouping

Add 4x^{2} to both sides.

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

Combine -6x^{2} and 4x^{2} to get -2x^{2}.

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

Subtract x from both sides.

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

Combine 9x and -x to get 8x.

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

Divide both sides by 2.

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

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

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

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.

$$a=3 b=1$$

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

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

Factor out -x in -x^{2}+3x.

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

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

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

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

$$x=3 x=1$$

Steps Using the Quadratic Formula

Add 4x^{2} to both sides.

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

Combine -6x^{2} and 4x^{2} to get -2x^{2}.

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

Subtract x from both sides.

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

Combine 9x and -x to get 8x.

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

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

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

Square 8.

$$x=\frac{-8±\sqrt{64-4\left(-2\right)\left(-6\right)}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-8±\sqrt{64+8\left(-6\right)}}{2\left(-2\right)}$$

Multiply 8 times -6.

$$x=\frac{-8±\sqrt{64-48}}{2\left(-2\right)}$$

Add 64 to -48.

$$x=\frac{-8±\sqrt{16}}{2\left(-2\right)}$$

Take the square root of 16.

$$x=\frac{-8±4}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{-8±4}{-4}$$

Now solve the equation x=\frac{-8±4}{-4} when ± is plus. Add -8 to 4.

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

Divide -4 by -4.

$$x=1$$

Now solve the equation x=\frac{-8±4}{-4} when ± is minus. Subtract 4 from -8.

$$x=-\frac{12}{-4}$$

Divide -12 by -4.

$$x=3$$

The equation is now solved.

$$x=1 x=3$$

Steps for Completing the Square

Add 4x^{2} to both sides.

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

Combine -6x^{2} and 4x^{2} to get -2x^{2}.

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

Subtract x from both sides.

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

Combine 9x and -x to get 8x.

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

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

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

Divide both sides by -2.

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

Dividing by -2 undoes the multiplication by -2.

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

Divide 8 by -2.

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

Divide 6 by -2.

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

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

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

Square -2.

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

Add -3 to 4.

$$x^{2}-4x+4=1$$

Factor x^{2}-4x+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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-2=1 x-2=-1$$

Add 2 to both sides of the equation.

$$x=3 x=1$$