Solve x^2-3x=-1+3x

$$x ^ { 2 } - 3 x = - 1 + 3 x$$

Solve for x

$$x=2\sqrt{2}+3\approx 5.828427125$$

$$x=3-2\sqrt{2}\approx 0.171572875$$

Steps Using the Quadratic Formula

Subtract -1 from both sides.

$$x^{2}-3x-\left(-1\right)=3x$$

The opposite of -1 is 1.

$$x^{2}-3x+1=3x$$

Subtract 3x from both sides.

$$x^{2}-3x+1-3x=0$$

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

$$x^{2}-6x+1=0$$

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

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

Square -6.

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

Add 36 to -4.

$$x=\frac{-\left(-6\right)±\sqrt{32}}{2}$$

Take the square root of 32.

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

The opposite of -6 is 6.

$$x=\frac{6±4\sqrt{2}}{2}$$

Now solve the equation x=\frac{6±4\sqrt{2}}{2} when ± is plus. Add 6 to 4\sqrt{2}.

$$x=\frac{4\sqrt{2}+6}{2}$$

Divide 6+4\sqrt{2} by 2.

$$x=2\sqrt{2}+3$$

Now solve the equation x=\frac{6±4\sqrt{2}}{2} when ± is minus. Subtract 4\sqrt{2} from 6.

$$x=\frac{6-4\sqrt{2}}{2}$$

Divide 6-4\sqrt{2} by 2.

$$x=3-2\sqrt{2}$$

The equation is now solved.

$$x=2\sqrt{2}+3 x=3-2\sqrt{2}$$

Steps for Completing the Square

Subtract 3x from both sides.

$$x^{2}-3x-3x=-1$$

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

$$x^{2}-6x=-1$$

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.

$$x^{2}-6x+\left(-3\right)^{2}=-1+\left(-3\right)^{2}$$

Square -3.

$$x^{2}-6x+9=-1+9$$

Add -1 to 9.

$$x^{2}-6x+9=8$$

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}.

$$\left(x-3\right)^{2}=8$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-3\right)^{2}}=\sqrt{8}$$

Simplify.

$$x-3=2\sqrt{2} x-3=-2\sqrt{2}$$

Add 3 to both sides of the equation.

$$x=2\sqrt{2}+3 x=3-2\sqrt{2}$$