Solve x=2piquad5x^2+2x+2=?

$$x = 2 \pi \quad 5 x ^ { 2 } + 2 x + 2 = ?$$

Solve for x (complex solution)

$$x=-\frac{1+i\sqrt{-\left(1-80\pi \right)}}{20\pi }\approx -0.015915494-0.251810791i$$

$$x=-\frac{-i\sqrt{-\left(1-80\pi \right)}+1}{20\pi }\approx -0.015915494+0.251810791i$$

Steps Using the Quadratic Formula

Multiply 2 and 5 to get 10.

$$x=10\pi x^{2}+2x+2$$

Subtract 10\pi x^{2} from both sides.

$$x-10\pi x^{2}=2x+2$$

Subtract 2x from both sides.

$$x-10\pi x^{2}-2x=2$$

Subtract 2 from both sides.

$$x-10\pi x^{2}-2x-2=0$$

Combine x and -2x to get -x.

$$-x-10\pi x^{2}-2=0$$

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.

$$\left(-10\pi \right)x^{2}-x-2=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -10\pi for a, -1 for b, and -2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Multiply -4 times -10\pi .

$$x=\frac{-\left(-1\right)±\sqrt{1+40\pi \left(-2\right)}}{2\left(-10\pi \right)}$$

Multiply 40\pi times -2.

$$x=\frac{-\left(-1\right)±\sqrt{1-80\pi }}{2\left(-10\pi \right)}$$

Take the square root of 1-80\pi .

$$x=\frac{-\left(-1\right)±i\sqrt{-\left(1-80\pi \right)}}{2\left(-10\pi \right)}$$

The opposite of -1 is 1.

$$x=\frac{1±i\sqrt{-\left(1-80\pi \right)}}{2\left(-10\pi \right)}$$

Multiply 2 times -10\pi .

$$x=\frac{1±i\sqrt{-\left(1-80\pi \right)}}{-20\pi }$$

Now solve the equation x=\frac{1±i\sqrt{-\left(1-80\pi \right)}}{-20\pi } when ± is plus. Add 1 to i\sqrt{-\left(1-80\pi \right)}.

$$x=\frac{1+i\sqrt{80\pi -1}}{-20\pi }$$

Divide 1+i\sqrt{-1+80\pi } by -20\pi .

$$x=-\frac{1+i\sqrt{80\pi -1}}{20\pi }$$

Now solve the equation x=\frac{1±i\sqrt{-\left(1-80\pi \right)}}{-20\pi } when ± is minus. Subtract i\sqrt{-\left(1-80\pi \right)} from 1.

$$x=\frac{-i\sqrt{80\pi -1}+1}{-20\pi }$$

Divide 1-i\sqrt{-1+80\pi } by -20\pi .

$$x=-\frac{-i\sqrt{80\pi -1}+1}{20\pi }$$

The equation is now solved.

$$x=-\frac{1+i\sqrt{80\pi -1}}{20\pi } x=-\frac{-i\sqrt{80\pi -1}+1}{20\pi }$$

Steps for Completing the Square

Multiply 2 and 5 to get 10.

$$x=10\pi x^{2}+2x+2$$

Subtract 10\pi x^{2} from both sides.

$$x-10\pi x^{2}=2x+2$$

Subtract 2x from both sides.

$$x-10\pi x^{2}-2x=2$$

Combine x and -2x to get -x.

$$-x-10\pi x^{2}=2$$

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.

$$\left(-10\pi \right)x^{2}-x=2$$

Divide both sides by -10\pi .

$$\frac{\left(-10\pi \right)x^{2}-x}{-10\pi }=\frac{2}{-10\pi }$$

Dividing by -10\pi undoes the multiplication by -10\pi .

$$x^{2}+\left(-\frac{1}{-10\pi }\right)x=\frac{2}{-10\pi }$$

Divide -1 by -10\pi .

$$x^{2}+\frac{1}{10\pi }x=\frac{2}{-10\pi }$$

Divide 2 by -10\pi .

$$x^{2}+\frac{1}{10\pi }x=-\frac{1}{5\pi }$$

Divide \frac{1}{10\pi }, the coefficient of the x term, by 2 to get \frac{1}{20\pi }. Then add the square of \frac{1}{20\pi } to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{1}{10\pi }x+\left(\frac{1}{20\pi }\right)^{2}=-\frac{1}{5\pi }+\left(\frac{1}{20\pi }\right)^{2}$$

Square \frac{1}{20\pi }.

$$x^{2}+\frac{1}{10\pi }x+\frac{1}{400\pi ^{2}}=-\frac{1}{5\pi }+\frac{1}{400\pi ^{2}}$$

Add -\frac{1}{5\pi } to \frac{1}{400\pi ^{2}}.

$$x^{2}+\frac{1}{10\pi }x+\frac{1}{400\pi ^{2}}=\frac{-\frac{\pi }{5}+\frac{1}{400}}{\pi ^{2}}$$

Factor x^{2}+\frac{1}{10\pi }x+\frac{1}{400\pi ^{2}}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{20\pi }\right)^{2}=\frac{-\frac{\pi }{5}+\frac{1}{400}}{\pi ^{2}}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{20\pi }\right)^{2}}=\sqrt{\frac{-\frac{\pi }{5}+\frac{1}{400}}{\pi ^{2}}}$$

Simplify.

$$x+\frac{1}{20\pi }=\frac{i\sqrt{-\left(1-80\pi \right)}}{20\pi } x+\frac{1}{20\pi }=-\frac{i\sqrt{80\pi -1}}{20\pi }$$

Subtract \frac{1}{20\pi } from both sides of the equation.

$$x=\frac{-1+i\sqrt{-\left(1-80\pi \right)}}{20\pi } x=-\frac{1+i\sqrt{80\pi -1}}{20\pi }$$