Solve 3+4q^2=8q

$$3 + 4 q ^ { 2 } = 8 q$$

Solve for q

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

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

Steps Using Factoring By Grouping

Subtract 8q from both sides.

$$3+4q^{2}-8q=0$$

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

$$4q^{2}-8q+3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 4q^{2}+aq+bq+3. To find a and b, set up a system to be solved.

$$a+b=-8 ab=4\times 3=12$$

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 12.

$$-1,-12 -2,-6 -3,-4$$

Calculate the sum for each pair.

$$-1-12=-13 -2-6=-8 -3-4=-7$$

The solution is the pair that gives sum -8.

$$a=-6 b=-2$$

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

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

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

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

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

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

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

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

Steps Using the Quadratic Formula

Subtract 8q from both sides.

$$3+4q^{2}-8q=0$$

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.

$$4q^{2}-8q+3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -8 for b, and 3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$q=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 4\times 3}}{2\times 4}$$

Square -8.

$$q=\frac{-\left(-8\right)±\sqrt{64-4\times 4\times 3}}{2\times 4}$$

Multiply -4 times 4.

$$q=\frac{-\left(-8\right)±\sqrt{64-16\times 3}}{2\times 4}$$

Multiply -16 times 3.

$$q=\frac{-\left(-8\right)±\sqrt{64-48}}{2\times 4}$$

Add 64 to -48.

$$q=\frac{-\left(-8\right)±\sqrt{16}}{2\times 4}$$

Take the square root of 16.

$$q=\frac{-\left(-8\right)±4}{2\times 4}$$

The opposite of -8 is 8.

$$q=\frac{8±4}{2\times 4}$$

Multiply 2 times 4.

$$q=\frac{8±4}{8}$$

Now solve the equation q=\frac{8±4}{8} when ± is plus. Add 8 to 4.

$$q=\frac{12}{8}$$

Reduce the fraction \frac{12}{8} to lowest terms by extracting and canceling out 4.

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

Now solve the equation q=\frac{8±4}{8} when ± is minus. Subtract 4 from 8.

$$q=\frac{4}{8}$$

Reduce the fraction \frac{4}{8} to lowest terms by extracting and canceling out 4.

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

The equation is now solved.

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

Steps for Completing the Square

Subtract 8q from both sides.

$$3+4q^{2}-8q=0$$

Subtract 3 from both sides. Anything subtracted from zero gives its negation.

$$4q^{2}-8q=-3$$

Divide both sides by 4.

$$\frac{4q^{2}-8q}{4}=-\frac{3}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$q^{2}+\left(-\frac{8}{4}\right)q=-\frac{3}{4}$$

Divide -8 by 4.

$$q^{2}-2q=-\frac{3}{4}$$

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

$$q^{2}-2q+1=-\frac{3}{4}+1$$

Add -\frac{3}{4} to 1.

$$q^{2}-2q+1=\frac{1}{4}$$

Factor q^{2}-2q+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Add 1 to both sides of the equation.

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