Solve 2x^2+9x=x-368

$$2 x ^ { 2 } + 9 x = x - 368$$

Solve for x (complex solution)

$$x=-2+6\sqrt{5}i\approx -2+13.416407865i$$

$$x=-6\sqrt{5}i-2\approx -2-13.416407865i$$

Steps Using the Quadratic Formula

Subtract x from both sides.

$$2x^{2}+9x-x=-368$$

Combine 9x and -x to get 8x.

$$2x^{2}+8x=-368$$

Add 368 to both sides.

$$2x^{2}+8x+368=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 8 for b, and 368 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-8±\sqrt{8^{2}-4\times 2\times 368}}{2\times 2}$$

Square 8.

$$x=\frac{-8±\sqrt{64-4\times 2\times 368}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-8±\sqrt{64-8\times 368}}{2\times 2}$$

Multiply -8 times 368.

$$x=\frac{-8±\sqrt{64-2944}}{2\times 2}$$

Add 64 to -2944.

$$x=\frac{-8±\sqrt{-2880}}{2\times 2}$$

Take the square root of -2880.

$$x=\frac{-8±24\sqrt{5}i}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-8±24\sqrt{5}i}{4}$$

Now solve the equation x=\frac{-8±24\sqrt{5}i}{4} when ± is plus. Add -8 to 24i\sqrt{5}.

$$x=\frac{-8+24\sqrt{5}i}{4}$$

Divide -8+24i\sqrt{5} by 4.

$$x=-2+6\sqrt{5}i$$

Now solve the equation x=\frac{-8±24\sqrt{5}i}{4} when ± is minus. Subtract 24i\sqrt{5} from -8.

$$x=\frac{-24\sqrt{5}i-8}{4}$$

Divide -8-24i\sqrt{5} by 4.

$$x=-6\sqrt{5}i-2$$

The equation is now solved.

$$x=-2+6\sqrt{5}i x=-6\sqrt{5}i-2$$

Steps for Completing the Square

Subtract x from both sides.

$$2x^{2}+9x-x=-368$$

Combine 9x and -x to get 8x.

$$2x^{2}+8x=-368$$

Divide both sides by 2.

$$\frac{2x^{2}+8x}{2}=-\frac{368}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{8}{2}x=-\frac{368}{2}$$

Divide 8 by 2.

$$x^{2}+4x=-\frac{368}{2}$$

Divide -368 by 2.

$$x^{2}+4x=-184$$

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

$$x^{2}+4x+2^{2}=-184+2^{2}$$

Square 2.

$$x^{2}+4x+4=-184+4$$

Add -184 to 4.

$$x^{2}+4x+4=-180$$

Factor x^{2}+4x+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}.

$$\left(x+2\right)^{2}=-180$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+2\right)^{2}}=\sqrt{-180}$$

Simplify.

$$x+2=6\sqrt{5}i x+2=-6\sqrt{5}i$$

Subtract 2 from both sides of the equation.

$$x=-2+6\sqrt{5}i x=-6\sqrt{5}i-2$$