Solve -x^2+1/2x+3/2=x+4

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

Solve for x (complex solution)

$$x=\frac{-1+\sqrt{39}i}{4}\approx -0.25+1.5612495i$$

$$x=\frac{-\sqrt{39}i-1}{4}\approx -0.25-1.5612495i$$

Steps Using the Quadratic Formula

Subtract x from both sides.

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

Combine \frac{1}{2}x and -x to get -\frac{1}{2}x.

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

Subtract 4 from both sides.

$$-x^{2}-\frac{1}{2}x+\frac{3}{2}-4=0$$

Subtract 4 from \frac{3}{2} to get -\frac{5}{2}.

$$-x^{2}-\frac{1}{2}x-\frac{5}{2}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -\frac{1}{2} for b, and -\frac{5}{2} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-\frac{1}{2}\right)±\sqrt{\left(-\frac{1}{2}\right)^{2}-4\left(-1\right)\left(-\frac{5}{2}\right)}}{2\left(-1\right)}$$

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

$$x=\frac{-\left(-\frac{1}{2}\right)±\sqrt{\frac{1}{4}-4\left(-1\right)\left(-\frac{5}{2}\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-\frac{1}{2}\right)±\sqrt{\frac{1}{4}+4\left(-\frac{5}{2}\right)}}{2\left(-1\right)}$$

Multiply 4 times -\frac{5}{2}.

$$x=\frac{-\left(-\frac{1}{2}\right)±\sqrt{\frac{1}{4}-10}}{2\left(-1\right)}$$

Add \frac{1}{4} to -10.

$$x=\frac{-\left(-\frac{1}{2}\right)±\sqrt{-\frac{39}{4}}}{2\left(-1\right)}$$

Take the square root of -\frac{39}{4}.

$$x=\frac{-\left(-\frac{1}{2}\right)±\frac{\sqrt{39}i}{2}}{2\left(-1\right)}$$

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

$$x=\frac{\frac{1}{2}±\frac{\sqrt{39}i}{2}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{\frac{1}{2}±\frac{\sqrt{39}i}{2}}{-2}$$

Now solve the equation x=\frac{\frac{1}{2}±\frac{\sqrt{39}i}{2}}{-2} when ± is plus. Add \frac{1}{2} to \frac{i\sqrt{39}}{2}.

$$x=\frac{1+\sqrt{39}i}{-2\times 2}$$

Divide \frac{1+i\sqrt{39}}{2} by -2.

$$x=\frac{-\sqrt{39}i-1}{4}$$

Now solve the equation x=\frac{\frac{1}{2}±\frac{\sqrt{39}i}{2}}{-2} when ± is minus. Subtract \frac{i\sqrt{39}}{2} from \frac{1}{2}.

$$x=\frac{-\sqrt{39}i+1}{-2\times 2}$$

Divide \frac{1-i\sqrt{39}}{2} by -2.

$$x=\frac{-1+\sqrt{39}i}{4}$$

The equation is now solved.

$$x=\frac{-\sqrt{39}i-1}{4} x=\frac{-1+\sqrt{39}i}{4}$$

Steps for Completing the Square

Subtract x from both sides.

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

Combine \frac{1}{2}x and -x to get -\frac{1}{2}x.

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

Subtract \frac{3}{2} from both sides.

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

Subtract \frac{3}{2} from 4 to get \frac{5}{2}.

$$-x^{2}-\frac{1}{2}x=\frac{5}{2}$$

Divide both sides by -1.

$$\frac{-x^{2}-\frac{1}{2}x}{-1}=\frac{\frac{5}{2}}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\left(-\frac{\frac{1}{2}}{-1}\right)x=\frac{\frac{5}{2}}{-1}$$

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

$$x^{2}+\frac{1}{2}x=\frac{\frac{5}{2}}{-1}$$

Divide \frac{5}{2} by -1.

$$x^{2}+\frac{1}{2}x=-\frac{5}{2}$$

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.

$$x^{2}+\frac{1}{2}x+\left(\frac{1}{4}\right)^{2}=-\frac{5}{2}+\left(\frac{1}{4}\right)^{2}$$

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

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=-\frac{5}{2}+\frac{1}{16}$$

Add -\frac{5}{2} to \frac{1}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=-\frac{39}{16}$$

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}.

$$\left(x+\frac{1}{4}\right)^{2}=-\frac{39}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{4}\right)^{2}}=\sqrt{-\frac{39}{16}}$$

Simplify.

$$x+\frac{1}{4}=\frac{\sqrt{39}i}{4} x+\frac{1}{4}=-\frac{\sqrt{39}i}{4}$$

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

$$x=\frac{-1+\sqrt{39}i}{4} x=\frac{-\sqrt{39}i-1}{4}$$