Solve x^2+x+3=x+21x^2

$$x ^ { 2 } + x + 3 = x + 21 x ^ { 2 }$$

Solve for x

$$x=\frac{\sqrt{15}}{10}\approx 0.387298335$$

$$x=-\frac{\sqrt{15}}{10}\approx -0.387298335$$

Steps by Finding Square Root

Subtract x from both sides.

$$x^{2}+x+3-x=21x^{2}$$

Combine x and -x to get 0.

$$x^{2}+3=21x^{2}$$

Subtract 21x^{2} from both sides.

$$x^{2}+3-21x^{2}=0$$

Combine x^{2} and -21x^{2} to get -20x^{2}.

$$-20x^{2}+3=0$$

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

$$-20x^{2}=-3$$

Divide both sides by -20.

$$x^{2}=\frac{-3}{-20}$$

Fraction \frac{-3}{-20} can be simplified to \frac{3}{20} by removing the negative sign from both the numerator and the denominator.

$$x^{2}=\frac{3}{20}$$

Take the square root of both sides of the equation.

$$x=\frac{\sqrt{15}}{10} x=-\frac{\sqrt{15}}{10}$$

Steps Using the Quadratic Formula

Subtract x from both sides.

$$x^{2}+x+3-x=21x^{2}$$

Combine x and -x to get 0.

$$x^{2}+3=21x^{2}$$

Subtract 21x^{2} from both sides.

$$x^{2}+3-21x^{2}=0$$

Combine x^{2} and -21x^{2} to get -20x^{2}.

$$-20x^{2}+3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -20 for a, 0 for b, and 3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-20\right)\times 3}}{2\left(-20\right)}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-20\right)\times 3}}{2\left(-20\right)}$$

Multiply -4 times -20.

$$x=\frac{0±\sqrt{80\times 3}}{2\left(-20\right)}$$

Multiply 80 times 3.

$$x=\frac{0±\sqrt{240}}{2\left(-20\right)}$$

Take the square root of 240.

$$x=\frac{0±4\sqrt{15}}{2\left(-20\right)}$$

Multiply 2 times -20.

$$x=\frac{0±4\sqrt{15}}{-40}$$

Now solve the equation x=\frac{0±4\sqrt{15}}{-40} when ± is plus.

$$x=-\frac{\sqrt{15}}{10}$$

Now solve the equation x=\frac{0±4\sqrt{15}}{-40} when ± is minus.

$$x=\frac{\sqrt{15}}{10}$$

The equation is now solved.

$$x=-\frac{\sqrt{15}}{10} x=\frac{\sqrt{15}}{10}$$