Solve x^2+(15/2-x)^2=(9-x-1/2)

$$x ^ { 2 } + ( \frac { 15 } { 2 } - x ) ^ { 2 } = ( 9 - x - \frac { 1 } { 2 } )$$

Solve for x (complex solution)

$$x=\frac{\sqrt{186}i}{4}+\frac{7}{2}\approx 3.5+3.409545424i$$

$$x=-\frac{\sqrt{186}i}{4}+\frac{7}{2}\approx 3.5-3.409545424i$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\frac{15}{2}-x\right)^{2}.

$$x^{2}+\frac{225}{4}-15x+x^{2}=9-x-\frac{1}{2}$$

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

$$2x^{2}+\frac{225}{4}-15x=9-x-\frac{1}{2}$$

Subtract \frac{1}{2} from 9 to get \frac{17}{2}.

$$2x^{2}+\frac{225}{4}-15x=\frac{17}{2}-x$$

Subtract \frac{17}{2} from both sides.

$$2x^{2}+\frac{225}{4}-15x-\frac{17}{2}=-x$$

Subtract \frac{17}{2} from \frac{225}{4} to get \frac{191}{4}.

$$2x^{2}+\frac{191}{4}-15x=-x$$

Add x to both sides.

$$2x^{2}+\frac{191}{4}-15x+x=0$$

Combine -15x and x to get -14x.

$$2x^{2}+\frac{191}{4}-14x=0$$

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.

$$2x^{2}-14x+\frac{191}{4}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -14 for b, and \frac{191}{4} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-14\right)±\sqrt{\left(-14\right)^{2}-4\times 2\times \frac{191}{4}}}{2\times 2}$$

Square -14.

$$x=\frac{-\left(-14\right)±\sqrt{196-4\times 2\times \frac{191}{4}}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-14\right)±\sqrt{196-8\times \frac{191}{4}}}{2\times 2}$$

Multiply -8 times \frac{191}{4}.

$$x=\frac{-\left(-14\right)±\sqrt{196-382}}{2\times 2}$$

Add 196 to -382.

$$x=\frac{-\left(-14\right)±\sqrt{-186}}{2\times 2}$$

Take the square root of -186.

$$x=\frac{-\left(-14\right)±\sqrt{186}i}{2\times 2}$$

The opposite of -14 is 14.

$$x=\frac{14±\sqrt{186}i}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{14±\sqrt{186}i}{4}$$

Now solve the equation x=\frac{14±\sqrt{186}i}{4} when ± is plus. Add 14 to i\sqrt{186}.

$$x=\frac{14+\sqrt{186}i}{4}$$

Divide 14+i\sqrt{186} by 4.

$$x=\frac{\sqrt{186}i}{4}+\frac{7}{2}$$

Now solve the equation x=\frac{14±\sqrt{186}i}{4} when ± is minus. Subtract i\sqrt{186} from 14.

$$x=\frac{-\sqrt{186}i+14}{4}$$

Divide 14-i\sqrt{186} by 4.

$$x=-\frac{\sqrt{186}i}{4}+\frac{7}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{186}i}{4}+\frac{7}{2} x=-\frac{\sqrt{186}i}{4}+\frac{7}{2}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\frac{15}{2}-x\right)^{2}.

$$x^{2}+\frac{225}{4}-15x+x^{2}=9-x-\frac{1}{2}$$

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

$$2x^{2}+\frac{225}{4}-15x=9-x-\frac{1}{2}$$

Subtract \frac{1}{2} from 9 to get \frac{17}{2}.

$$2x^{2}+\frac{225}{4}-15x=\frac{17}{2}-x$$

Add x to both sides.

$$2x^{2}+\frac{225}{4}-15x+x=\frac{17}{2}$$

Combine -15x and x to get -14x.

$$2x^{2}+\frac{225}{4}-14x=\frac{17}{2}$$

Subtract \frac{225}{4} from both sides.

$$2x^{2}-14x=\frac{17}{2}-\frac{225}{4}$$

Subtract \frac{225}{4} from \frac{17}{2} to get -\frac{191}{4}.

$$2x^{2}-14x=-\frac{191}{4}$$

Divide both sides by 2.

$$\frac{2x^{2}-14x}{2}=-\frac{\frac{191}{4}}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\left(-\frac{14}{2}\right)x=-\frac{\frac{191}{4}}{2}$$

Divide -14 by 2.

$$x^{2}-7x=-\frac{\frac{191}{4}}{2}$$

Divide -\frac{191}{4} by 2.

$$x^{2}-7x=-\frac{191}{8}$$

Divide -7, the coefficient of the x term, by 2 to get -\frac{7}{2}. Then add the square of -\frac{7}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-7x+\left(-\frac{7}{2}\right)^{2}=-\frac{191}{8}+\left(-\frac{7}{2}\right)^{2}$$

Square -\frac{7}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-7x+\frac{49}{4}=-\frac{191}{8}+\frac{49}{4}$$

Add -\frac{191}{8} to \frac{49}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-7x+\frac{49}{4}=-\frac{93}{8}$$

Factor x^{2}-7x+\frac{49}{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}.

$$\left(x-\frac{7}{2}\right)^{2}=-\frac{93}{8}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{2}\right)^{2}}=\sqrt{-\frac{93}{8}}$$

Simplify.

$$x-\frac{7}{2}=\frac{\sqrt{186}i}{4} x-\frac{7}{2}=-\frac{\sqrt{186}i}{4}$$

Add \frac{7}{2} to both sides of the equation.

$$x=\frac{\sqrt{186}i}{4}+\frac{7}{2} x=-\frac{\sqrt{186}i}{4}+\frac{7}{2}$$