Solve -x=-10-3x^2

$$- x = - 10 - 3 x ^ { 2 }$$

Solve for x (complex solution)

$$x=\frac{1+\sqrt{119}i}{6}\approx 0.166666667+1.818118686i$$

$$x=\frac{-\sqrt{119}i+1}{6}\approx 0.166666667-1.818118686i$$

Steps Using the Quadratic Formula

Subtract -10 from both sides.

$$-x-\left(-10\right)=-3x^{2}$$

The opposite of -10 is 10.

$$-x+10=-3x^{2}$$

Add 3x^{2} to both sides.

$$-x+10+3x^{2}=0$$

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.

$$3x^{2}-x+10=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -1 for b, and 10 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\times 3\times 10}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-1\right)±\sqrt{1-12\times 10}}{2\times 3}$$

Multiply -12 times 10.

$$x=\frac{-\left(-1\right)±\sqrt{1-120}}{2\times 3}$$

Add 1 to -120.

$$x=\frac{-\left(-1\right)±\sqrt{-119}}{2\times 3}$$

Take the square root of -119.

$$x=\frac{-\left(-1\right)±\sqrt{119}i}{2\times 3}$$

The opposite of -1 is 1.

$$x=\frac{1±\sqrt{119}i}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{1±\sqrt{119}i}{6}$$

Now solve the equation x=\frac{1±\sqrt{119}i}{6} when ± is plus. Add 1 to i\sqrt{119}.

$$x=\frac{1+\sqrt{119}i}{6}$$

Now solve the equation x=\frac{1±\sqrt{119}i}{6} when ± is minus. Subtract i\sqrt{119} from 1.

$$x=\frac{-\sqrt{119}i+1}{6}$$

The equation is now solved.

$$x=\frac{1+\sqrt{119}i}{6} x=\frac{-\sqrt{119}i+1}{6}$$

Steps for Completing the Square

Add 3x^{2} to both sides.

$$-x+3x^{2}=-10$$

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.

$$3x^{2}-x=-10$$

Divide both sides by 3.

$$\frac{3x^{2}-x}{3}=-\frac{10}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}-\frac{1}{3}x=-\frac{10}{3}$$

Divide -\frac{1}{3}, the coefficient of the x term, by 2 to get -\frac{1}{6}. Then add the square of -\frac{1}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1}{3}x+\left(-\frac{1}{6}\right)^{2}=-\frac{10}{3}+\left(-\frac{1}{6}\right)^{2}$$

Square -\frac{1}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{3}x+\frac{1}{36}=-\frac{10}{3}+\frac{1}{36}$$

Add -\frac{10}{3} to \frac{1}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{1}{3}x+\frac{1}{36}=-\frac{119}{36}$$

Factor x^{2}-\frac{1}{3}x+\frac{1}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{6}\right)^{2}=-\frac{119}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{6}\right)^{2}}=\sqrt{-\frac{119}{36}}$$

Simplify.

$$x-\frac{1}{6}=\frac{\sqrt{119}i}{6} x-\frac{1}{6}=-\frac{\sqrt{119}i}{6}$$

Add \frac{1}{6} to both sides of the equation.

$$x=\frac{1+\sqrt{119}i}{6} x=\frac{-\sqrt{119}i+1}{6}$$