Solve x=-1/2x^2+3x+7/2

$$x=- \frac{ 1 }{ 2 } { x }^{ 2 } +3x+ \frac{ 7 }{ 2 }$$

Solve for x

$$x=\sqrt{11}+2\approx 5.31662479$$

$$x=2-\sqrt{11}\approx -1.31662479$$

Steps Using the Quadratic Formula

Add \frac{1}{2}x^{2} to both sides.

$$x+\frac{1}{2}x^{2}=3x+\frac{7}{2}$$

Subtract 3x from both sides.

$$x+\frac{1}{2}x^{2}-3x=\frac{7}{2}$$

Combine x and -3x to get -2x.

$$-2x+\frac{1}{2}x^{2}=\frac{7}{2}$$

Subtract \frac{7}{2} from both sides.

$$-2x+\frac{1}{2}x^{2}-\frac{7}{2}=0$$

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.

$$\frac{1}{2}x^{2}-2x-\frac{7}{2}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{1}{2} for a, -2 for b, and -\frac{7}{2} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times \frac{1}{2}\left(-\frac{7}{2}\right)}}{2\times \frac{1}{2}}$$

Square -2.

$$x=\frac{-\left(-2\right)±\sqrt{4-4\times \frac{1}{2}\left(-\frac{7}{2}\right)}}{2\times \frac{1}{2}}$$

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

$$x=\frac{-\left(-2\right)±\sqrt{4-2\left(-\frac{7}{2}\right)}}{2\times \frac{1}{2}}$$

Multiply -2 times -\frac{7}{2}.

$$x=\frac{-\left(-2\right)±\sqrt{4+7}}{2\times \frac{1}{2}}$$

Add 4 to 7.

$$x=\frac{-\left(-2\right)±\sqrt{11}}{2\times \frac{1}{2}}$$

The opposite of -2 is 2.

$$x=\frac{2±\sqrt{11}}{2\times \frac{1}{2}}$$

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

$$x=\frac{2±\sqrt{11}}{1}$$

Now solve the equation x=\frac{2±\sqrt{11}}{1} when ± is plus. Add 2 to \sqrt{11}.

$$x=\frac{\sqrt{11}+2}{1}$$

Divide 2+\sqrt{11} by 1.

$$x=\sqrt{11}+2$$

Now solve the equation x=\frac{2±\sqrt{11}}{1} when ± is minus. Subtract \sqrt{11} from 2.

$$x=\frac{2-\sqrt{11}}{1}$$

Divide 2-\sqrt{11} by 1.

$$x=2-\sqrt{11}$$

The equation is now solved.

$$x=\sqrt{11}+2 x=2-\sqrt{11}$$

Steps for Completing the Square

Add \frac{1}{2}x^{2} to both sides.

$$x+\frac{1}{2}x^{2}=3x+\frac{7}{2}$$

Subtract 3x from both sides.

$$x+\frac{1}{2}x^{2}-3x=\frac{7}{2}$$

Combine x and -3x to get -2x.

$$-2x+\frac{1}{2}x^{2}=\frac{7}{2}$$

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.

$$\frac{1}{2}x^{2}-2x=\frac{7}{2}$$

Multiply both sides by 2.

$$\frac{\frac{1}{2}x^{2}-2x}{\frac{1}{2}}=\frac{\frac{7}{2}}{\frac{1}{2}}$$

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

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

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

$$x^{2}-4x=\frac{\frac{7}{2}}{\frac{1}{2}}$$

Divide \frac{7}{2} by \frac{1}{2} by multiplying \frac{7}{2} by the reciprocal of \frac{1}{2}.

$$x^{2}-4x=7$$

Divide -4, the coefficient of the x term, by 2 to get -2. Then add the square of -2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-4x+\left(-2\right)^{2}=7+\left(-2\right)^{2}$$

Square -2.

$$x^{2}-4x+4=7+4$$

Add 7 to 4.

$$x^{2}-4x+4=11$$

Factor x^{2}-4x+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}.

$$\left(x-2\right)^{2}=11$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-2\right)^{2}}=\sqrt{11}$$

Simplify.

$$x-2=\sqrt{11} x-2=-\sqrt{11}$$

Add 2 to both sides of the equation.

$$x=\sqrt{11}+2 x=2-\sqrt{11}$$