Solve x^2+45=x-2

$${ x }^{ 2 } +45=x-2$$

Solve for x (complex solution)

$$x=\frac{1+\sqrt{187}i}{2}\approx 0.5+6.837397166i$$

$$x=\frac{-\sqrt{187}i+1}{2}\approx 0.5-6.837397166i$$

Steps Using the Quadratic Formula

Subtract x from both sides.

$$x^{2}+45-x=-2$$

Add 2 to both sides.

$$x^{2}+45-x+2=0$$

Add 45 and 2 to get 47.

$$x^{2}+47-x=0$$

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.

$$x^{2}-x+47=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -1 for b, and 47 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\times 47}}{2}$$

Multiply -4 times 47.

$$x=\frac{-\left(-1\right)±\sqrt{1-188}}{2}$$

Add 1 to -188.

$$x=\frac{-\left(-1\right)±\sqrt{-187}}{2}$$

Take the square root of -187.

$$x=\frac{-\left(-1\right)±\sqrt{187}i}{2}$$

The opposite of -1 is 1.

$$x=\frac{1±\sqrt{187}i}{2}$$

Now solve the equation x=\frac{1±\sqrt{187}i}{2} when ± is plus. Add 1 to i\sqrt{187}.

$$x=\frac{1+\sqrt{187}i}{2}$$

Now solve the equation x=\frac{1±\sqrt{187}i}{2} when ± is minus. Subtract i\sqrt{187} from 1.

$$x=\frac{-\sqrt{187}i+1}{2}$$

The equation is now solved.

$$x=\frac{1+\sqrt{187}i}{2} x=\frac{-\sqrt{187}i+1}{2}$$

Steps for Completing the Square

Subtract x from both sides.

$$x^{2}+45-x=-2$$

Subtract 45 from both sides.

$$x^{2}-x=-2-45$$

Subtract 45 from -2 to get -47.

$$x^{2}-x=-47$$

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

$$x^{2}-x+\left(-\frac{1}{2}\right)^{2}=-47+\left(-\frac{1}{2}\right)^{2}$$

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

$$x^{2}-x+\frac{1}{4}=-47+\frac{1}{4}$$

Add -47 to \frac{1}{4}.

$$x^{2}-x+\frac{1}{4}=-\frac{187}{4}$$

Factor x^{2}-x+\frac{1}{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}.

$$\left(x-\frac{1}{2}\right)^{2}=-\frac{187}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{-\frac{187}{4}}$$

Simplify.

$$x-\frac{1}{2}=\frac{\sqrt{187}i}{2} x-\frac{1}{2}=-\frac{\sqrt{187}i}{2}$$

Add \frac{1}{2} to both sides of the equation.

$$x=\frac{1+\sqrt{187}i}{2} x=\frac{-\sqrt{187}i+1}{2}$$