Solve 2x^2-5x+10=x(6-x)

$$2 x ^ { 2 } - 5 x + 10 = x ( 6 - x )$$

Solve for x

$$x=2$$

$$x = \frac{5}{3} = 1\frac{2}{3} \approx 1.666666667$$

Steps Using Factoring By Grouping

Use the distributive property to multiply x by 6-x.

$$2x^{2}-5x+10=6x-x^{2}$$

Subtract 6x from both sides.

$$2x^{2}-5x+10-6x=-x^{2}$$

Combine -5x and -6x to get -11x.

$$2x^{2}-11x+10=-x^{2}$$

Add x^{2} to both sides.

$$2x^{2}-11x+10+x^{2}=0$$

Combine 2x^{2} and x^{2} to get 3x^{2}.

$$3x^{2}-11x+10=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3x^{2}+ax+bx+10. To find a and b, set up a system to be solved.

$$a+b=-11 ab=3\times 10=30$$

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 30.

$$-1,-30 -2,-15 -3,-10 -5,-6$$

Calculate the sum for each pair.

$$-1-30=-31 -2-15=-17 -3-10=-13 -5-6=-11$$

The solution is the pair that gives sum -11.

$$a=-6 b=-5$$

Rewrite 3x^{2}-11x+10 as \left(3x^{2}-6x\right)+\left(-5x+10\right).

$$\left(3x^{2}-6x\right)+\left(-5x+10\right)$$

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

$$3x\left(x-2\right)-5\left(x-2\right)$$

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

$$\left(x-2\right)\left(3x-5\right)$$

To find equation solutions, solve x-2=0 and 3x-5=0.

$$x=2 x=\frac{5}{3}$$

Steps Using the Quadratic Formula

Use the distributive property to multiply x by 6-x.

$$2x^{2}-5x+10=6x-x^{2}$$

Subtract 6x from both sides.

$$2x^{2}-5x+10-6x=-x^{2}$$

Combine -5x and -6x to get -11x.

$$2x^{2}-11x+10=-x^{2}$$

Add x^{2} to both sides.

$$2x^{2}-11x+10+x^{2}=0$$

Combine 2x^{2} and x^{2} to get 3x^{2}.

$$3x^{2}-11x+10=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -11 for b, and 10 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\times 3\times 10}}{2\times 3}$$

Square -11.

$$x=\frac{-\left(-11\right)±\sqrt{121-4\times 3\times 10}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-11\right)±\sqrt{121-12\times 10}}{2\times 3}$$

Multiply -12 times 10.

$$x=\frac{-\left(-11\right)±\sqrt{121-120}}{2\times 3}$$

Add 121 to -120.

$$x=\frac{-\left(-11\right)±\sqrt{1}}{2\times 3}$$

Take the square root of 1.

$$x=\frac{-\left(-11\right)±1}{2\times 3}$$

The opposite of -11 is 11.

$$x=\frac{11±1}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{11±1}{6}$$

Now solve the equation x=\frac{11±1}{6} when ± is plus. Add 11 to 1.

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

Divide 12 by 6.

$$x=2$$

Now solve the equation x=\frac{11±1}{6} when ± is minus. Subtract 1 from 11.

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

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

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

The equation is now solved.

$$x=2 x=\frac{5}{3}$$

Steps for Completing the Square

Use the distributive property to multiply x by 6-x.

$$2x^{2}-5x+10=6x-x^{2}$$

Subtract 6x from both sides.

$$2x^{2}-5x+10-6x=-x^{2}$$

Combine -5x and -6x to get -11x.

$$2x^{2}-11x+10=-x^{2}$$

Add x^{2} to both sides.

$$2x^{2}-11x+10+x^{2}=0$$

Combine 2x^{2} and x^{2} to get 3x^{2}.

$$3x^{2}-11x+10=0$$

Subtract 10 from both sides. Anything subtracted from zero gives its negation.

$$3x^{2}-11x=-10$$

Divide both sides by 3.

$$\frac{3x^{2}-11x}{3}=-\frac{10}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}-\frac{11}{3}x=-\frac{10}{3}$$

Divide -\frac{11}{3}, the coefficient of the x term, by 2 to get -\frac{11}{6}. Then add the square of -\frac{11}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{11}{3}x+\left(-\frac{11}{6}\right)^{2}=-\frac{10}{3}+\left(-\frac{11}{6}\right)^{2}$$

Square -\frac{11}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{11}{3}x+\frac{121}{36}=-\frac{10}{3}+\frac{121}{36}$$

Add -\frac{10}{3} to \frac{121}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{11}{3}x+\frac{121}{36}=\frac{1}{36}$$

Factor x^{2}-\frac{11}{3}x+\frac{121}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{11}{6}\right)^{2}=\frac{1}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{11}{6}\right)^{2}}=\sqrt{\frac{1}{36}}$$

Simplify.

$$x-\frac{11}{6}=\frac{1}{6} x-\frac{11}{6}=-\frac{1}{6}$$

Add \frac{11}{6} to both sides of the equation.

$$x=2 x=\frac{5}{3}$$