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

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

Solve for x

$$x=2$$

$$x=\frac{2}{3}\approx 0.666666667$$

Steps Using Factoring By Grouping

Combine -4x and -6x to get -10x.

$$6x-4=6x^{2}-10x+4$$

Subtract 6x^{2} from both sides.

$$6x-4-6x^{2}=-10x+4$$

Add 10x to both sides.

$$6x-4-6x^{2}+10x=4$$

Combine 6x and 10x to get 16x.

$$16x-4-6x^{2}=4$$

Subtract 4 from both sides.

$$16x-4-6x^{2}-4=0$$

Subtract 4 from -4 to get -8.

$$16x-8-6x^{2}=0$$

Divide both sides by 2.

$$8x-4-3x^{2}=0$$

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

$$-3x^{2}+8x-4=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -3x^{2}+ax+bx-4. To find a and b, set up a system to be solved.

$$a+b=8 ab=-3\left(-4\right)=12$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 12.

$$1,12 2,6 3,4$$

Calculate the sum for each pair.

$$1+12=13 2+6=8 3+4=7$$

The solution is the pair that gives sum 8.

$$a=6 b=2$$

Rewrite -3x^{2}+8x-4 as \left(-3x^{2}+6x\right)+\left(2x-4\right).

$$\left(-3x^{2}+6x\right)+\left(2x-4\right)$$

Factor out 3x in the first and -2 in the second group.

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

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

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

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

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

Steps Using the Quadratic Formula

Combine -4x and -6x to get -10x.

$$6x-4=6x^{2}-10x+4$$

Subtract 6x^{2} from both sides.

$$6x-4-6x^{2}=-10x+4$$

Add 10x to both sides.

$$6x-4-6x^{2}+10x=4$$

Combine 6x and 10x to get 16x.

$$16x-4-6x^{2}=4$$

Subtract 4 from both sides.

$$16x-4-6x^{2}-4=0$$

Subtract 4 from -4 to get -8.

$$16x-8-6x^{2}=0$$

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.

$$-6x^{2}+16x-8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -6 for a, 16 for b, and -8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-16±\sqrt{16^{2}-4\left(-6\right)\left(-8\right)}}{2\left(-6\right)}$$

Square 16.

$$x=\frac{-16±\sqrt{256-4\left(-6\right)\left(-8\right)}}{2\left(-6\right)}$$

Multiply -4 times -6.

$$x=\frac{-16±\sqrt{256+24\left(-8\right)}}{2\left(-6\right)}$$

Multiply 24 times -8.

$$x=\frac{-16±\sqrt{256-192}}{2\left(-6\right)}$$

Add 256 to -192.

$$x=\frac{-16±\sqrt{64}}{2\left(-6\right)}$$

Take the square root of 64.

$$x=\frac{-16±8}{2\left(-6\right)}$$

Multiply 2 times -6.

$$x=\frac{-16±8}{-12}$$

Now solve the equation x=\frac{-16±8}{-12} when ± is plus. Add -16 to 8.

$$x=-\frac{8}{-12}$$

Reduce the fraction \frac{-8}{-12} to lowest terms by extracting and canceling out 4.

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

Now solve the equation x=\frac{-16±8}{-12} when ± is minus. Subtract 8 from -16.

$$x=-\frac{24}{-12}$$

Divide -24 by -12.

$$x=2$$

The equation is now solved.

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

Steps for Completing the Square

Combine -4x and -6x to get -10x.

$$6x-4=6x^{2}-10x+4$$

Subtract 6x^{2} from both sides.

$$6x-4-6x^{2}=-10x+4$$

Add 10x to both sides.

$$6x-4-6x^{2}+10x=4$$

Combine 6x and 10x to get 16x.

$$16x-4-6x^{2}=4$$

Add 4 to both sides.

$$16x-6x^{2}=4+4$$

Add 4 and 4 to get 8.

$$16x-6x^{2}=8$$

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.

$$-6x^{2}+16x=8$$

Divide both sides by -6.

$$\frac{-6x^{2}+16x}{-6}=\frac{8}{-6}$$

Dividing by -6 undoes the multiplication by -6.

$$x^{2}+\frac{16}{-6}x=\frac{8}{-6}$$

Reduce the fraction \frac{16}{-6} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{8}{3}x=\frac{8}{-6}$$

Reduce the fraction \frac{8}{-6} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{8}{3}x=-\frac{4}{3}$$

Divide -\frac{8}{3}, the coefficient of the x term, by 2 to get -\frac{4}{3}. Then add the square of -\frac{4}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{8}{3}x+\left(-\frac{4}{3}\right)^{2}=-\frac{4}{3}+\left(-\frac{4}{3}\right)^{2}$$

Square -\frac{4}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{8}{3}x+\frac{16}{9}=-\frac{4}{3}+\frac{16}{9}$$

Add -\frac{4}{3} to \frac{16}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{8}{3}x+\frac{16}{9}=\frac{4}{9}$$

Factor x^{2}-\frac{8}{3}x+\frac{16}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-\frac{4}{3}=\frac{2}{3} x-\frac{4}{3}=-\frac{2}{3}$$

Add \frac{4}{3} to both sides of the equation.

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