Solve q+q^2=3q^2-3

$$q + q ^ { 2 } = 3 q ^ { 2 } - 3$$

Solve for q

$$q=-1$$

$$q = \frac{3}{2} = 1\frac{1}{2} = 1.5$$

Steps Using Factoring By Grouping

Subtract 3q^{2} from both sides.

$$q+q^{2}-3q^{2}=-3$$

Combine q^{2} and -3q^{2} to get -2q^{2}.

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

Add 3 to both sides.

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

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

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

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

$$a+b=1 ab=-2\times 3=-6$$

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.

$$a=3 b=-2$$

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

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

Factor out -q in the first and -1 in the second group.

$$-q\left(2q-3\right)-\left(2q-3\right)$$

Factor out common term 2q-3 by using distributive property.

$$\left(2q-3\right)\left(-q-1\right)$$

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

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

Steps Using the Quadratic Formula

Subtract 3q^{2} from both sides.

$$q+q^{2}-3q^{2}=-3$$

Combine q^{2} and -3q^{2} to get -2q^{2}.

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

Add 3 to both sides.

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

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.

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

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}.

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

Square 1.

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

Multiply -4 times -2.

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

Multiply 8 times 3.

$$q=\frac{-1±\sqrt{1+24}}{2\left(-2\right)}$$

Add 1 to 24.

$$q=\frac{-1±\sqrt{25}}{2\left(-2\right)}$$

Take the square root of 25.

$$q=\frac{-1±5}{2\left(-2\right)}$$

Multiply 2 times -2.

$$q=\frac{-1±5}{-4}$$

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

$$q=\frac{4}{-4}$$

Divide 4 by -4.

$$q=-1$$

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

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

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

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

The equation is now solved.

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

Steps for Completing the Square

Subtract 3q^{2} from both sides.

$$q+q^{2}-3q^{2}=-3$$

Combine q^{2} and -3q^{2} to get -2q^{2}.

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

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.

$$-2q^{2}+q=-3$$

Divide both sides by -2.

$$\frac{-2q^{2}+q}{-2}=-\frac{3}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$q^{2}+\frac{1}{-2}q=-\frac{3}{-2}$$

Divide 1 by -2.

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

Divide -3 by -2.

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

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.

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

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

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

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.

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

Factor q^{2}-\frac{1}{2}q+\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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$q-\frac{1}{4}=\frac{5}{4} q-\frac{1}{4}=-\frac{5}{4}$$

Add \frac{1}{4} to both sides of the equation.

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