Solve 3x^2=7x-10

$$3 x ^ { 2 } = 7 x - 10$$

Solve for x (complex solution)

$$x=\frac{7+\sqrt{71}i}{6}\approx 1.166666667+1.404358296i$$

$$x=\frac{-\sqrt{71}i+7}{6}\approx 1.166666667-1.404358296i$$

Steps Using the Quadratic Formula

Subtract 7x from both sides.

$$3x^{2}-7x=-10$$

Add 10 to both sides.

$$3x^{2}-7x+10=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -7 for b, and 10 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 3\times 10}}{2\times 3}$$

Square -7.

$$x=\frac{-\left(-7\right)±\sqrt{49-4\times 3\times 10}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-7\right)±\sqrt{49-12\times 10}}{2\times 3}$$

Multiply -12 times 10.

$$x=\frac{-\left(-7\right)±\sqrt{49-120}}{2\times 3}$$

Add 49 to -120.

$$x=\frac{-\left(-7\right)±\sqrt{-71}}{2\times 3}$$

Take the square root of -71.

$$x=\frac{-\left(-7\right)±\sqrt{71}i}{2\times 3}$$

The opposite of -7 is 7.

$$x=\frac{7±\sqrt{71}i}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{7±\sqrt{71}i}{6}$$

Now solve the equation x=\frac{7±\sqrt{71}i}{6} when ± is plus. Add 7 to i\sqrt{71}.

$$x=\frac{7+\sqrt{71}i}{6}$$

Now solve the equation x=\frac{7±\sqrt{71}i}{6} when ± is minus. Subtract i\sqrt{71} from 7.

$$x=\frac{-\sqrt{71}i+7}{6}$$

The equation is now solved.

$$x=\frac{7+\sqrt{71}i}{6} x=\frac{-\sqrt{71}i+7}{6}$$

Steps for Completing the Square

Subtract 7x from both sides.

$$3x^{2}-7x=-10$$

Divide both sides by 3.

$$\frac{3x^{2}-7x}{3}=-\frac{10}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}-\frac{7}{3}x=-\frac{10}{3}$$

Divide -\frac{7}{3}, the coefficient of the x term, by 2 to get -\frac{7}{6}. Then add the square of -\frac{7}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{7}{3}x+\left(-\frac{7}{6}\right)^{2}=-\frac{10}{3}+\left(-\frac{7}{6}\right)^{2}$$

Square -\frac{7}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{7}{3}x+\frac{49}{36}=-\frac{10}{3}+\frac{49}{36}$$

Add -\frac{10}{3} to \frac{49}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{7}{3}x+\frac{49}{36}=-\frac{71}{36}$$

Factor x^{2}-\frac{7}{3}x+\frac{49}{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}.

$$\left(x-\frac{7}{6}\right)^{2}=-\frac{71}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{6}\right)^{2}}=\sqrt{-\frac{71}{36}}$$

Simplify.

$$x-\frac{7}{6}=\frac{\sqrt{71}i}{6} x-\frac{7}{6}=-\frac{\sqrt{71}i}{6}$$

Add \frac{7}{6} to both sides of the equation.

$$x=\frac{7+\sqrt{71}i}{6} x=\frac{-\sqrt{71}i+7}{6}$$