Solve 4x^2+8=-5x

$$4 x ^ { 2 } + 8 = - 5 x$$

Solve for x (complex solution)

$$x=\frac{-5+\sqrt{103}i}{8}\approx -0.625+1.268611446i$$

$$x=\frac{-\sqrt{103}i-5}{8}\approx -0.625-1.268611446i$$

Steps Using the Quadratic Formula

Add 5x to both sides.

$$4x^{2}+8+5x=0$$

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.

$$4x^{2}+5x+8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, 5 for b, and 8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-5±\sqrt{5^{2}-4\times 4\times 8}}{2\times 4}$$

Square 5.

$$x=\frac{-5±\sqrt{25-4\times 4\times 8}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-5±\sqrt{25-16\times 8}}{2\times 4}$$

Multiply -16 times 8.

$$x=\frac{-5±\sqrt{25-128}}{2\times 4}$$

Add 25 to -128.

$$x=\frac{-5±\sqrt{-103}}{2\times 4}$$

Take the square root of -103.

$$x=\frac{-5±\sqrt{103}i}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-5±\sqrt{103}i}{8}$$

Now solve the equation x=\frac{-5±\sqrt{103}i}{8} when ± is plus. Add -5 to i\sqrt{103}.

$$x=\frac{-5+\sqrt{103}i}{8}$$

Now solve the equation x=\frac{-5±\sqrt{103}i}{8} when ± is minus. Subtract i\sqrt{103} from -5.

$$x=\frac{-\sqrt{103}i-5}{8}$$

The equation is now solved.

$$x=\frac{-5+\sqrt{103}i}{8} x=\frac{-\sqrt{103}i-5}{8}$$

Steps for Completing the Square

Add 5x to both sides.

$$4x^{2}+8+5x=0$$

Subtract 8 from both sides. Anything subtracted from zero gives its negation.

$$4x^{2}+5x=-8$$

Divide both sides by 4.

$$\frac{4x^{2}+5x}{4}=-\frac{8}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\frac{5}{4}x=-\frac{8}{4}$$

Divide -8 by 4.

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

Divide \frac{5}{4}, the coefficient of the x term, by 2 to get \frac{5}{8}. Then add the square of \frac{5}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{5}{4}x+\left(\frac{5}{8}\right)^{2}=-2+\left(\frac{5}{8}\right)^{2}$$

Square \frac{5}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{5}{4}x+\frac{25}{64}=-2+\frac{25}{64}$$

Add -2 to \frac{25}{64}.

$$x^{2}+\frac{5}{4}x+\frac{25}{64}=-\frac{103}{64}$$

Factor x^{2}+\frac{5}{4}x+\frac{25}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{5}{8}\right)^{2}=-\frac{103}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{8}\right)^{2}}=\sqrt{-\frac{103}{64}}$$

Simplify.

$$x+\frac{5}{8}=\frac{\sqrt{103}i}{8} x+\frac{5}{8}=-\frac{\sqrt{103}i}{8}$$

Subtract \frac{5}{8} from both sides of the equation.

$$x=\frac{-5+\sqrt{103}i}{8} x=\frac{-\sqrt{103}i-5}{8}$$