Solve -x^2+9x-3=3x

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

Solve for x

$$x=\sqrt{6}+3\approx 5.449489743$$

$$x=3-\sqrt{6}\approx 0.550510257$$

Steps Using the Quadratic Formula

Subtract 3x from both sides.

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

Combine 9x and -3x to get 6x.

$$-x^{2}+6x-3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 6 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-6±\sqrt{6^{2}-4\left(-1\right)\left(-3\right)}}{2\left(-1\right)}$$

Square 6.

$$x=\frac{-6±\sqrt{36-4\left(-1\right)\left(-3\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-6±\sqrt{36+4\left(-3\right)}}{2\left(-1\right)}$$

Multiply 4 times -3.

$$x=\frac{-6±\sqrt{36-12}}{2\left(-1\right)}$$

Add 36 to -12.

$$x=\frac{-6±\sqrt{24}}{2\left(-1\right)}$$

Take the square root of 24.

$$x=\frac{-6±2\sqrt{6}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-6±2\sqrt{6}}{-2}$$

Now solve the equation x=\frac{-6±2\sqrt{6}}{-2} when ± is plus. Add -6 to 2\sqrt{6}.

$$x=\frac{2\sqrt{6}-6}{-2}$$

Divide -6+2\sqrt{6} by -2.

$$x=3-\sqrt{6}$$

Now solve the equation x=\frac{-6±2\sqrt{6}}{-2} when ± is minus. Subtract 2\sqrt{6} from -6.

$$x=\frac{-2\sqrt{6}-6}{-2}$$

Divide -6-2\sqrt{6} by -2.

$$x=\sqrt{6}+3$$

The equation is now solved.

$$x=3-\sqrt{6} x=\sqrt{6}+3$$

Steps for Completing the Square

Subtract 3x from both sides.

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

Combine 9x and -3x to get 6x.

$$-x^{2}+6x-3=0$$

Add 3 to both sides. Anything plus zero gives itself.

$$-x^{2}+6x=3$$

Divide both sides by -1.

$$\frac{-x^{2}+6x}{-1}=\frac{3}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{6}{-1}x=\frac{3}{-1}$$

Divide 6 by -1.

$$x^{2}-6x=\frac{3}{-1}$$

Divide 3 by -1.

$$x^{2}-6x=-3$$

Divide -6, the coefficient of the x term, by 2 to get -3. Then add the square of -3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-6x+\left(-3\right)^{2}=-3+\left(-3\right)^{2}$$

Square -3.

$$x^{2}-6x+9=-3+9$$

Add -3 to 9.

$$x^{2}-6x+9=6$$

Factor x^{2}-6x+9. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-3\right)^{2}=6$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-3\right)^{2}}=\sqrt{6}$$

Simplify.

$$x-3=\sqrt{6} x-3=-\sqrt{6}$$

Add 3 to both sides of the equation.

$$x=\sqrt{6}+3 x=3-\sqrt{6}$$