Solve 8x^2+6=-4x(3-2

$$8 x ^ { 2 } + 6 = - 4 x ( 3 - 2$$

Solve for x (complex solution)

$$x=\frac{-1+\sqrt{11}i}{4}\approx -0.25+0.829156198i$$

$$x=\frac{-\sqrt{11}i-1}{4}\approx -0.25-0.829156198i$$

Steps Using the Quadratic Formula

Subtract 2 from 3 to get 1.

$$8x^{2}+6=-4x$$

Add 4x to both sides.

$$8x^{2}+6+4x=0$$

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.

$$8x^{2}+4x+6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 8 for a, 4 for b, and 6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-4±\sqrt{4^{2}-4\times 8\times 6}}{2\times 8}$$

Square 4.

$$x=\frac{-4±\sqrt{16-4\times 8\times 6}}{2\times 8}$$

Multiply -4 times 8.

$$x=\frac{-4±\sqrt{16-32\times 6}}{2\times 8}$$

Multiply -32 times 6.

$$x=\frac{-4±\sqrt{16-192}}{2\times 8}$$

Add 16 to -192.

$$x=\frac{-4±\sqrt{-176}}{2\times 8}$$

Take the square root of -176.

$$x=\frac{-4±4\sqrt{11}i}{2\times 8}$$

Multiply 2 times 8.

$$x=\frac{-4±4\sqrt{11}i}{16}$$

Now solve the equation x=\frac{-4±4\sqrt{11}i}{16} when ± is plus. Add -4 to 4i\sqrt{11}.

$$x=\frac{-4+4\sqrt{11}i}{16}$$

Divide -4+4i\sqrt{11} by 16.

$$x=\frac{-1+\sqrt{11}i}{4}$$

Now solve the equation x=\frac{-4±4\sqrt{11}i}{16} when ± is minus. Subtract 4i\sqrt{11} from -4.

$$x=\frac{-4\sqrt{11}i-4}{16}$$

Divide -4-4i\sqrt{11} by 16.

$$x=\frac{-\sqrt{11}i-1}{4}$$

The equation is now solved.

$$x=\frac{-1+\sqrt{11}i}{4} x=\frac{-\sqrt{11}i-1}{4}$$

Steps for Completing the Square

Subtract 2 from 3 to get 1.

$$8x^{2}+6=-4x$$

Add 4x to both sides.

$$8x^{2}+6+4x=0$$

Subtract 6 from both sides. Anything subtracted from zero gives its negation.

$$8x^{2}+4x=-6$$

Divide both sides by 8.

$$\frac{8x^{2}+4x}{8}=-\frac{6}{8}$$

Dividing by 8 undoes the multiplication by 8.

$$x^{2}+\frac{4}{8}x=-\frac{6}{8}$$

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

$$x^{2}+\frac{1}{2}x=-\frac{6}{8}$$

Reduce the fraction \frac{-6}{8} to lowest terms by extracting and canceling out 2.

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

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.

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

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

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=-\frac{3}{4}+\frac{1}{16}$$

Add -\frac{3}{4} to \frac{1}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=-\frac{11}{16}$$

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}.

$$\left(x+\frac{1}{4}\right)^{2}=-\frac{11}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{4}\right)^{2}}=\sqrt{-\frac{11}{16}}$$

Simplify.

$$x+\frac{1}{4}=\frac{\sqrt{11}i}{4} x+\frac{1}{4}=-\frac{\sqrt{11}i}{4}$$

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

$$x=\frac{-1+\sqrt{11}i}{4} x=\frac{-\sqrt{11}i-1}{4}$$