Solve -6x^2=-20x+16

$$- 6 x ^ { 2 } = - 20 x + 16$$

Solve for x

$$x=2$$

$$x = \frac{4}{3} = 1\frac{1}{3} \approx 1.333333333$$

Steps Using Factoring By Grouping

Add 20x to both sides.

$$-6x^{2}+20x=16$$

Subtract 16 from both sides.

$$-6x^{2}+20x-16=0$$

Divide both sides by 2.

$$-3x^{2}+10x-8=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -3x^{2}+ax+bx-8. To find a and b, set up a system to be solved.

$$a+b=10 ab=-3\left(-8\right)=24$$

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 24.

$$1,24 2,12 3,8 4,6$$

Calculate the sum for each pair.

$$1+24=25 2+12=14 3+8=11 4+6=10$$

The solution is the pair that gives sum 10.

$$a=6 b=4$$

Rewrite -3x^{2}+10x-8 as \left(-3x^{2}+6x\right)+\left(4x-8\right).

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

Factor out 3x in the first and -4 in the second group.

$$3x\left(-x+2\right)-4\left(-x+2\right)$$

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

$$\left(-x+2\right)\left(3x-4\right)$$

To find equation solutions, solve -x+2=0 and 3x-4=0.

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

Steps Using the Quadratic Formula

Add 20x to both sides.

$$-6x^{2}+20x=16$$

Subtract 16 from both sides.

$$-6x^{2}+20x-16=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -6 for a, 20 for b, and -16 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-20±\sqrt{20^{2}-4\left(-6\right)\left(-16\right)}}{2\left(-6\right)}$$

Square 20.

$$x=\frac{-20±\sqrt{400-4\left(-6\right)\left(-16\right)}}{2\left(-6\right)}$$

Multiply -4 times -6.

$$x=\frac{-20±\sqrt{400+24\left(-16\right)}}{2\left(-6\right)}$$

Multiply 24 times -16.

$$x=\frac{-20±\sqrt{400-384}}{2\left(-6\right)}$$

Add 400 to -384.

$$x=\frac{-20±\sqrt{16}}{2\left(-6\right)}$$

Take the square root of 16.

$$x=\frac{-20±4}{2\left(-6\right)}$$

Multiply 2 times -6.

$$x=\frac{-20±4}{-12}$$

Now solve the equation x=\frac{-20±4}{-12} when ± is plus. Add -20 to 4.

$$x=-\frac{16}{-12}$$

Reduce the fraction \frac{-16}{-12} to lowest terms by extracting and canceling out 4.

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

Now solve the equation x=\frac{-20±4}{-12} when ± is minus. Subtract 4 from -20.

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

Divide -24 by -12.

$$x=2$$

The equation is now solved.

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

Steps for Completing the Square

Add 20x to both sides.

$$-6x^{2}+20x=16$$

Divide both sides by -6.

$$\frac{-6x^{2}+20x}{-6}=\frac{16}{-6}$$

Dividing by -6 undoes the multiplication by -6.

$$x^{2}+\frac{20}{-6}x=\frac{16}{-6}$$

Reduce the fraction \frac{20}{-6} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{10}{3}x=\frac{16}{-6}$$

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

$$x^{2}-\frac{10}{3}x=-\frac{8}{3}$$

Divide -\frac{10}{3}, the coefficient of the x term, by 2 to get -\frac{5}{3}. Then add the square of -\frac{5}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{10}{3}x+\left(-\frac{5}{3}\right)^{2}=-\frac{8}{3}+\left(-\frac{5}{3}\right)^{2}$$

Square -\frac{5}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{10}{3}x+\frac{25}{9}=-\frac{8}{3}+\frac{25}{9}$$

Add -\frac{8}{3} to \frac{25}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{10}{3}x+\frac{25}{9}=\frac{1}{9}$$

Factor x^{2}-\frac{10}{3}x+\frac{25}{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}.

$$\left(x-\frac{5}{3}\right)^{2}=\frac{1}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{3}\right)^{2}}=\sqrt{\frac{1}{9}}$$

Simplify.

$$x-\frac{5}{3}=\frac{1}{3} x-\frac{5}{3}=-\frac{1}{3}$$

Add \frac{5}{3} to both sides of the equation.

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