Solve 4x^2=-4x-17

$$4 { x }^{ 2 } =-4x-17$$

Solve for x (complex solution)

$$x=-\frac{1}{2}+2i=-0.5+2i$$

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

Steps Using the Quadratic Formula

Add 4x to both sides.

$$4x^{2}+4x=-17$$

Add 17 to both sides.

$$4x^{2}+4x+17=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, 4 for b, and 17 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-4±\sqrt{4^{2}-4\times 4\times 17}}{2\times 4}$$

Square 4.

$$x=\frac{-4±\sqrt{16-4\times 4\times 17}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-4±\sqrt{16-16\times 17}}{2\times 4}$$

Multiply -16 times 17.

$$x=\frac{-4±\sqrt{16-272}}{2\times 4}$$

Add 16 to -272.

$$x=\frac{-4±\sqrt{-256}}{2\times 4}$$

Take the square root of -256.

$$x=\frac{-4±16i}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-4±16i}{8}$$

Now solve the equation x=\frac{-4±16i}{8} when ± is plus. Add -4 to 16i.

$$x=\frac{-4+16i}{8}$$

Divide -4+16i by 8.

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

Now solve the equation x=\frac{-4±16i}{8} when ± is minus. Subtract 16i from -4.

$$x=\frac{-4-16i}{8}$$

Divide -4-16i by 8.

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

The equation is now solved.

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

Steps for Completing the Square

Add 4x to both sides.

$$4x^{2}+4x=-17$$

Divide both sides by 4.

$$\frac{4x^{2}+4x}{4}=-\frac{17}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\frac{4}{4}x=-\frac{17}{4}$$

Divide 4 by 4.

$$x^{2}+x=-\frac{17}{4}$$

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

$$x^{2}+x+\left(\frac{1}{2}\right)^{2}=-\frac{17}{4}+\left(\frac{1}{2}\right)^{2}$$

Square \frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+x+\frac{1}{4}=\frac{-17+1}{4}$$

Add -\frac{17}{4} to \frac{1}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

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

Factor x^{2}+x+\frac{1}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Subtract \frac{1}{2} from both sides of the equation.

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