Solve 4x^2=8x-9

$$4 { x }^{ 2 } =8x-9$$

Solve for x (complex solution)

$$x=\frac{\sqrt{5}i}{2}+1\approx 1+1.118033989i$$

$$x=-\frac{\sqrt{5}i}{2}+1\approx 1-1.118033989i$$

Steps Using the Quadratic Formula

Subtract 8x from both sides.

$$4x^{2}-8x=-9$$

Add 9 to both sides.

$$4x^{2}-8x+9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -8 for b, and 9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 4\times 9}}{2\times 4}$$

Square -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\times 4\times 9}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-8\right)±\sqrt{64-16\times 9}}{2\times 4}$$

Multiply -16 times 9.

$$x=\frac{-\left(-8\right)±\sqrt{64-144}}{2\times 4}$$

Add 64 to -144.

$$x=\frac{-\left(-8\right)±\sqrt{-80}}{2\times 4}$$

Take the square root of -80.

$$x=\frac{-\left(-8\right)±4\sqrt{5}i}{2\times 4}$$

The opposite of -8 is 8.

$$x=\frac{8±4\sqrt{5}i}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{8±4\sqrt{5}i}{8}$$

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

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

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

$$x=\frac{\sqrt{5}i}{2}+1$$

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

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

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

$$x=-\frac{\sqrt{5}i}{2}+1$$

The equation is now solved.

$$x=\frac{\sqrt{5}i}{2}+1 x=-\frac{\sqrt{5}i}{2}+1$$

Steps for Completing the Square

Subtract 8x from both sides.

$$4x^{2}-8x=-9$$

Divide both sides by 4.

$$\frac{4x^{2}-8x}{4}=-\frac{9}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\left(-\frac{8}{4}\right)x=-\frac{9}{4}$$

Divide -8 by 4.

$$x^{2}-2x=-\frac{9}{4}$$

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.

$$x^{2}-2x+1=-\frac{9}{4}+1$$

Add -\frac{9}{4} to 1.

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

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

$$\left(x-1\right)^{2}=-\frac{5}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-1\right)^{2}}=\sqrt{-\frac{5}{4}}$$

Simplify.

$$x-1=\frac{\sqrt{5}i}{2} x-1=-\frac{\sqrt{5}i}{2}$$

Add 1 to both sides of the equation.

$$x=\frac{\sqrt{5}i}{2}+1 x=-\frac{\sqrt{5}i}{2}+1$$