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

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

Solve for x

$$x=-2$$

$$x = \frac{3}{2} = 1\frac{1}{2} = 1.5$$

Steps Using the Quadratic Formula

Subtract x^{2} from both sides.

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

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.

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

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

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

Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.

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

Multiply -4 times -1.

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

Multiply 4 times 3.

$$x=\frac{-\left(-\frac{1}{2}\right)±\sqrt{\frac{1}{4}+12}}{2\left(-1\right)}$$

Add \frac{1}{4} to 12.

$$x=\frac{-\left(-\frac{1}{2}\right)±\sqrt{\frac{49}{4}}}{2\left(-1\right)}$$

Take the square root of \frac{49}{4}.

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

The opposite of -\frac{1}{2} is \frac{1}{2}.

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

Multiply 2 times -1.

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

Now solve the equation x=\frac{\frac{1}{2}±\frac{7}{2}}{-2} when ± is plus. Add \frac{1}{2} to \frac{7}{2} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

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

Divide 4 by -2.

$$x=-2$$

Now solve the equation x=\frac{\frac{1}{2}±\frac{7}{2}}{-2} when ± is minus. Subtract \frac{7}{2} from \frac{1}{2} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

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

Divide -3 by -2.

$$x=\frac{3}{2}$$

The equation is now solved.

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

Steps for Completing the Square

Subtract x^{2} from both sides.

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

Subtract 3 from both sides. Anything subtracted from zero gives its negation.

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

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.

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

Divide both sides by -1.

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

Dividing by -1 undoes the multiplication by -1.

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

Divide -\frac{1}{2} by -1.

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

Divide -3 by -1.

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

Divide \frac{1}{2}, the coefficient of the x term, by 2 to get \frac{1}{4}. Then add the square of \frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{1}{2}x+\left(\frac{1}{4}\right)^{2}=3+\left(\frac{1}{4}\right)^{2}$$

Square \frac{1}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=3+\frac{1}{16}$$

Add 3 to \frac{1}{16}.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{49}{16}$$

Factor x^{2}+\frac{1}{2}x+\frac{1}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{4}\right)^{2}=\frac{49}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{4}\right)^{2}}=\sqrt{\frac{49}{16}}$$

Simplify.

$$x+\frac{1}{4}=\frac{7}{4} x+\frac{1}{4}=-\frac{7}{4}$$

Subtract \frac{1}{4} from both sides of the equation.

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