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

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

Solve for x (complex solution)

$$x=\sqrt{7}-1\approx 1.645751311$$

$$x=-\left(\sqrt{7}+1\right)\approx -3.645751311$$

Steps Using the Quadratic Formula

Subtract \frac{1}{2}x^{2} from both sides.

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

Subtract 2x from both sides.

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

Combine x and -2x to get -x.

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

Add 3 to both sides.

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

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.

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

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

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

Multiply -4 times -\frac{1}{2}.

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

Multiply 2 times 3.

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

Add 1 to 6.

$$x=\frac{-\left(-1\right)±\sqrt{7}}{2\left(-\frac{1}{2}\right)}$$

The opposite of -1 is 1.

$$x=\frac{1±\sqrt{7}}{2\left(-\frac{1}{2}\right)}$$

Multiply 2 times -\frac{1}{2}.

$$x=\frac{1±\sqrt{7}}{-1}$$

Now solve the equation x=\frac{1±\sqrt{7}}{-1} when ± is plus. Add 1 to \sqrt{7}.

$$x=\frac{\sqrt{7}+1}{-1}$$

Divide 1+\sqrt{7} by -1.

$$x=-\left(\sqrt{7}+1\right)$$

Now solve the equation x=\frac{1±\sqrt{7}}{-1} when ± is minus. Subtract \sqrt{7} from 1.

$$x=\frac{1-\sqrt{7}}{-1}$$

Divide 1-\sqrt{7} by -1.

$$x=\sqrt{7}-1$$

The equation is now solved.

$$x=-\left(\sqrt{7}+1\right) x=\sqrt{7}-1$$

Steps for Completing the Square

Subtract \frac{1}{2}x^{2} from both sides.

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

Subtract 2x from both sides.

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

Combine x and -2x to get -x.

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

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

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

Multiply both sides by -2.

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

Dividing by -\frac{1}{2} undoes the multiplication by -\frac{1}{2}.

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

Divide -1 by -\frac{1}{2} by multiplying -1 by the reciprocal of -\frac{1}{2}.

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

Divide -3 by -\frac{1}{2} by multiplying -3 by the reciprocal of -\frac{1}{2}.

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

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.

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

Square 1.

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

Add 6 to 1.

$$x^{2}+2x+1=7$$

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}.

$$\left(x+1\right)^{2}=7$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{7}$$

Simplify.

$$x+1=\sqrt{7} x+1=-\sqrt{7}$$

Subtract 1 from both sides of the equation.

$$x=\sqrt{7}-1 x=-\sqrt{7}-1$$

Solve for x

$$x=\sqrt{7}-1\approx 1.645751311$$

$$x=-\sqrt{7}-1\approx -3.645751311$$

Steps Using the Quadratic Formula

Subtract \frac{1}{2}x^{2} from both sides.

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

Subtract 2x from both sides.

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

Combine x and -2x to get -x.

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

Add 3 to both sides.

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

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.

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

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

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

Multiply -4 times -\frac{1}{2}.

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

Multiply 2 times 3.

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

Add 1 to 6.

$$x=\frac{-\left(-1\right)±\sqrt{7}}{2\left(-\frac{1}{2}\right)}$$

The opposite of -1 is 1.

$$x=\frac{1±\sqrt{7}}{2\left(-\frac{1}{2}\right)}$$

Multiply 2 times -\frac{1}{2}.

$$x=\frac{1±\sqrt{7}}{-1}$$

Now solve the equation x=\frac{1±\sqrt{7}}{-1} when ± is plus. Add 1 to \sqrt{7}.

$$x=\frac{\sqrt{7}+1}{-1}$$

Divide 1+\sqrt{7} by -1.

$$x=-\left(\sqrt{7}+1\right)$$

Now solve the equation x=\frac{1±\sqrt{7}}{-1} when ± is minus. Subtract \sqrt{7} from 1.

$$x=\frac{1-\sqrt{7}}{-1}$$

Divide 1-\sqrt{7} by -1.

$$x=\sqrt{7}-1$$

The equation is now solved.

$$x=-\left(\sqrt{7}+1\right) x=\sqrt{7}-1$$

Steps for Completing the Square

Subtract \frac{1}{2}x^{2} from both sides.

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

Subtract 2x from both sides.

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

Combine x and -2x to get -x.

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

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

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

Multiply both sides by -2.

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

Dividing by -\frac{1}{2} undoes the multiplication by -\frac{1}{2}.

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

Divide -1 by -\frac{1}{2} by multiplying -1 by the reciprocal of -\frac{1}{2}.

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

Divide -3 by -\frac{1}{2} by multiplying -3 by the reciprocal of -\frac{1}{2}.

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

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.

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

Square 1.

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

Add 6 to 1.

$$x^{2}+2x+1=7$$

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}.

$$\left(x+1\right)^{2}=7$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{7}$$

Simplify.

$$x+1=\sqrt{7} x+1=-\sqrt{7}$$

Subtract 1 from both sides of the equation.

$$x=\sqrt{7}-1 x=-\sqrt{7}-1$$