Solve 8q(q-2)+10=0

$$8 q ( q - 2 ) + 10 = 0$$

Solve for q

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

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

Steps Using the Quadratic Formula

Use the distributive property to multiply 8q by q-2.

$$8q^{2}-16q+10=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 8 for a, -16 for b, and 10 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$q=\frac{-\left(-16\right)±\sqrt{\left(-16\right)^{2}-4\times 8\times 10}}{2\times 8}$$

Square -16.

$$q=\frac{-\left(-16\right)±\sqrt{256-4\times 8\times 10}}{2\times 8}$$

Multiply -4 times 8.

$$q=\frac{-\left(-16\right)±\sqrt{256-32\times 10}}{2\times 8}$$

Multiply -32 times 10.

$$q=\frac{-\left(-16\right)±\sqrt{256-320}}{2\times 8}$$

Add 256 to -320.

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

Take the square root of -64.

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

The opposite of -16 is 16.

$$q=\frac{16±8i}{2\times 8}$$

Multiply 2 times 8.

$$q=\frac{16±8i}{16}$$

Now solve the equation q=\frac{16±8i}{16} when ± is plus. Add 16 to 8i.

$$q=\frac{16+8i}{16}$$

Divide 16+8i by 16.

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

Now solve the equation q=\frac{16±8i}{16} when ± is minus. Subtract 8i from 16.

$$q=\frac{16-8i}{16}$$

Divide 16-8i by 16.

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

The equation is now solved.

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

Steps for Completing the Square

Use the distributive property to multiply 8q by q-2.

$$8q^{2}-16q+10=0$$

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

$$8q^{2}-16q=-10$$

Divide both sides by 8.

$$\frac{8q^{2}-16q}{8}=-\frac{10}{8}$$

Dividing by 8 undoes the multiplication by 8.

$$q^{2}+\left(-\frac{16}{8}\right)q=-\frac{10}{8}$$

Divide -16 by 8.

$$q^{2}-2q=-\frac{10}{8}$$

Reduce the fraction \frac{-10}{8} to lowest terms by extracting and canceling out 2.

$$q^{2}-2q=-\frac{5}{4}$$

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.

$$q^{2}-2q+1=-\frac{5}{4}+1$$

Add -\frac{5}{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.

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

Add 1 to both sides of the equation.

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