Solve x=3-2x^2

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

Solve for x

$$x = -\frac{3}{2} = -1\frac{1}{2} = -1.5$$

$$x=1$$

Steps Using Factoring By Grouping

Subtract 3 from both sides.

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

Add 2x^{2} to both sides.

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

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

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

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

$$a+b=1 ab=2\left(-3\right)=-6$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -6.

$$-1,6 -2,3$$

Calculate the sum for each pair.

$$-1+6=5 -2+3=1$$

The solution is the pair that gives sum 1.

$$a=-2 b=3$$

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

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

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

$$2x\left(x-1\right)+3\left(x-1\right)$$

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

$$\left(x-1\right)\left(2x+3\right)$$

To find equation solutions, solve x-1=0 and 2x+3=0.

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

Steps Using the Quadratic Formula

Subtract 3 from both sides.

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

Add 2x^{2} to both sides.

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

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.

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

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

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

Square 1.

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

Multiply -4 times 2.

$$x=\frac{-1±\sqrt{1-8\left(-3\right)}}{2\times 2}$$

Multiply -8 times -3.

$$x=\frac{-1±\sqrt{1+24}}{2\times 2}$$

Add 1 to 24.

$$x=\frac{-1±\sqrt{25}}{2\times 2}$$

Take the square root of 25.

$$x=\frac{-1±5}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-1±5}{4}$$

Now solve the equation x=\frac{-1±5}{4} when ± is plus. Add -1 to 5.

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

Divide 4 by 4.

$$x=1$$

Now solve the equation x=\frac{-1±5}{4} when ± is minus. Subtract 5 from -1.

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

Reduce the fraction \frac{-6}{4} to lowest terms by extracting and canceling out 2.

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

The equation is now solved.

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

Steps for Completing the Square

Add 2x^{2} to both sides.

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

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.

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

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

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

Divide \frac{1}{2}, the coefficient of the x term, by 2 to get \frac{1}{4}. Then add the square of \frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square \frac{1}{4} by squaring both the numerator and the denominator of the fraction.

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

Add \frac{3}{2} to \frac{1}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{25}{16}$$

Factor x^{2}+\frac{1}{2}x+\frac{1}{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}.

$$\left(x+\frac{1}{4}\right)^{2}=\frac{25}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{4}\right)^{2}}=\sqrt{\frac{25}{16}}$$

Simplify.

$$x+\frac{1}{4}=\frac{5}{4} x+\frac{1}{4}=-\frac{5}{4}$$

Subtract \frac{1}{4} from both sides of the equation.

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