Solve 0=0.3*x^2-3x+9

$$0=0.3 \times { x }^{ 2 } -3x+9$$

Solve for x (complex solution)

$$x=5+i\sqrt{5}\approx 5+2.236067977i$$

$$x=-i\sqrt{5}+5\approx 5-2.236067977i$$

Steps Using the Quadratic Formula

Swap sides so that all variable terms are on the left hand side.

$$0.3x^{2}-3x+9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 0.3 for a, -3 for b, and 9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 0.3\times 9}}{2\times 0.3}$$

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\times 0.3\times 9}}{2\times 0.3}$$

Multiply -4 times 0.3.

$$x=\frac{-\left(-3\right)±\sqrt{9-1.2\times 9}}{2\times 0.3}$$

Multiply -1.2 times 9.

$$x=\frac{-\left(-3\right)±\sqrt{9-10.8}}{2\times 0.3}$$

Add 9 to -10.8.

$$x=\frac{-\left(-3\right)±\sqrt{-1.8}}{2\times 0.3}$$

Take the square root of -1.8.

$$x=\frac{-\left(-3\right)±\frac{3\sqrt{5}i}{5}}{2\times 0.3}$$

The opposite of -3 is 3.

$$x=\frac{3±\frac{3\sqrt{5}i}{5}}{2\times 0.3}$$

Multiply 2 times 0.3.

$$x=\frac{3±\frac{3\sqrt{5}i}{5}}{0.6}$$

Now solve the equation x=\frac{3±\frac{3\sqrt{5}i}{5}}{0.6} when ± is plus. Add 3 to \frac{3i\sqrt{5}}{5}.

$$x=\frac{\frac{3\sqrt{5}i}{5}+3}{0.6}$$

Divide 3+\frac{3i\sqrt{5}}{5} by 0.6 by multiplying 3+\frac{3i\sqrt{5}}{5} by the reciprocal of 0.6.

$$x=5+\sqrt{5}i$$

Now solve the equation x=\frac{3±\frac{3\sqrt{5}i}{5}}{0.6} when ± is minus. Subtract \frac{3i\sqrt{5}}{5} from 3.

$$x=\frac{-\frac{3\sqrt{5}i}{5}+3}{0.6}$$

Divide 3-\frac{3i\sqrt{5}}{5} by 0.6 by multiplying 3-\frac{3i\sqrt{5}}{5} by the reciprocal of 0.6.

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

The equation is now solved.

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

Steps for Completing the Square

Swap sides so that all variable terms are on the left hand side.

$$0.3x^{2}-3x+9=0$$

Subtract 9 from both sides. Anything subtracted from zero gives its negation.

$$0.3x^{2}-3x=-9$$

Divide both sides of the equation by 0.3, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{0.3x^{2}-3x}{0.3}=-\frac{9}{0.3}$$

Dividing by 0.3 undoes the multiplication by 0.3.

$$x^{2}+\left(-\frac{3}{0.3}\right)x=-\frac{9}{0.3}$$

Divide -3 by 0.3 by multiplying -3 by the reciprocal of 0.3.

$$x^{2}-10x=-\frac{9}{0.3}$$

Divide -9 by 0.3 by multiplying -9 by the reciprocal of 0.3.

$$x^{2}-10x=-30$$

Divide -10, the coefficient of the x term, by 2 to get -5. Then add the square of -5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-10x+\left(-5\right)^{2}=-30+\left(-5\right)^{2}$$

Square -5.

$$x^{2}-10x+25=-30+25$$

Add -30 to 25.

$$x^{2}-10x+25=-5$$

Factor x^{2}-10x+25. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Add 5 to both sides of the equation.

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