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

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

Solve for x

$$x=\frac{\sqrt{5}-3}{2}\approx -0.381966011$$

$$x=\frac{-\sqrt{5}-3}{2}\approx -2.618033989$$

Steps Using the Quadratic Formula

Use the distributive property to multiply x by x-3.

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

Subtract x^{2} from both sides.

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

Combine 2x^{2} and -x^{2} to get x^{2}.

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

Add 3x to both sides.

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

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

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

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

$$x=\frac{-3±\sqrt{3^{2}-4}}{2}$$

Square 3.

$$x=\frac{-3±\sqrt{9-4}}{2}$$

Add 9 to -4.

$$x=\frac{-3±\sqrt{5}}{2}$$

Now solve the equation x=\frac{-3±\sqrt{5}}{2} when ± is plus. Add -3 to \sqrt{5}.

$$x=\frac{\sqrt{5}-3}{2}$$

Now solve the equation x=\frac{-3±\sqrt{5}}{2} when ± is minus. Subtract \sqrt{5} from -3.

$$x=\frac{-\sqrt{5}-3}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{5}-3}{2} x=\frac{-\sqrt{5}-3}{2}$$

Steps for Completing the Square

Use the distributive property to multiply x by x-3.

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

Subtract x^{2} from both sides.

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

Combine 2x^{2} and -x^{2} to get x^{2}.

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

Add 3x to both sides.

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

Subtract 1 from both sides. Anything subtracted from zero gives its negation.

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

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

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

Square \frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+3x+\frac{9}{4}=-1+\frac{9}{4}$$

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

$$x^{2}+3x+\frac{9}{4}=\frac{5}{4}$$

Factor x^{2}+3x+\frac{9}{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}.

$$\left(x+\frac{3}{2}\right)^{2}=\frac{5}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{2}\right)^{2}}=\sqrt{\frac{5}{4}}$$

Simplify.

$$x+\frac{3}{2}=\frac{\sqrt{5}}{2} x+\frac{3}{2}=-\frac{\sqrt{5}}{2}$$

Subtract \frac{3}{2} from both sides of the equation.

$$x=\frac{\sqrt{5}-3}{2} x=\frac{-\sqrt{5}-3}{2}$$