Solve 2x^2-6x+10=-3x+9

$$2 x ^ { 2 } - 6 x + 10 = - 3 x + 9$$

Solve for x

$$x=\frac{1}{2}=0.5$$

$$x=1$$

Steps Using Factoring By Grouping

Add 3x to both sides.

$$2x^{2}-6x+10+3x=9$$

Combine -6x and 3x to get -3x.

$$2x^{2}-3x+10=9$$

Subtract 9 from both sides.

$$2x^{2}-3x+10-9=0$$

Subtract 9 from 10 to get 1.

$$2x^{2}-3x+1=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2x^{2}+ax+bx+1. To find a and b, set up a system to be solved.

$$a+b=-3 ab=2\times 1=2$$

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

$$a=-2 b=-1$$

Rewrite 2x^{2}-3x+1 as \left(2x^{2}-2x\right)+\left(-x+1\right).

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

Factor out 2x in the first and -1 in the second group.

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

Factor out common term x-1 by using distributive property.

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

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

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

Steps Using the Quadratic Formula

Add 3x to both sides.

$$2x^{2}-6x+10+3x=9$$

Combine -6x and 3x to get -3x.

$$2x^{2}-3x+10=9$$

Subtract 9 from both sides.

$$2x^{2}-3x+10-9=0$$

Subtract 9 from 10 to get 1.

$$2x^{2}-3x+1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -3 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square -3.

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

Multiply -4 times 2.

$$x=\frac{-\left(-3\right)±\sqrt{9-8}}{2\times 2}$$

Add 9 to -8.

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

Take the square root of 1.

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

The opposite of -3 is 3.

$$x=\frac{3±1}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{3±1}{4}$$

Now solve the equation x=\frac{3±1}{4} when ± is plus. Add 3 to 1.

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

Divide 4 by 4.

$$x=1$$

Now solve the equation x=\frac{3±1}{4} when ± is minus. Subtract 1 from 3.

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

Reduce the fraction \frac{2}{4} to lowest terms by extracting and canceling out 2.

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

The equation is now solved.

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

Steps for Completing the Square

Add 3x to both sides.

$$2x^{2}-6x+10+3x=9$$

Combine -6x and 3x to get -3x.

$$2x^{2}-3x+10=9$$

Subtract 10 from both sides.

$$2x^{2}-3x=9-10$$

Subtract 10 from 9 to get -1.

$$2x^{2}-3x=-1$$

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

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

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

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

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

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=-\frac{1}{2}+\frac{9}{16}$$

Add -\frac{1}{2} to \frac{9}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=\frac{1}{16}$$

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

$$\left(x-\frac{3}{4}\right)^{2}=\frac{1}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{4}\right)^{2}}=\sqrt{\frac{1}{16}}$$

Simplify.

$$x-\frac{3}{4}=\frac{1}{4} x-\frac{3}{4}=-\frac{1}{4}$$

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

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