Solve x^2-3x+10/x-4=0

$$\frac { x ^ { 2 } - 3 x + 10 } { x - 4 } = 0$$

Solve for x (complex solution)

$$x=\frac{3+\sqrt{31}i}{2}\approx 1.5+2.783882181i$$

$$x=\frac{-\sqrt{31}i+3}{2}\approx 1.5-2.783882181i$$

Steps Using the Quadratic Formula

Variable x cannot be equal to 4 since division by zero is not defined. Multiply both sides of the equation by x-4.

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

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

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

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\times 10}}{2}$$

Multiply -4 times 10.

$$x=\frac{-\left(-3\right)±\sqrt{9-40}}{2}$$

Add 9 to -40.

$$x=\frac{-\left(-3\right)±\sqrt{-31}}{2}$$

Take the square root of -31.

$$x=\frac{-\left(-3\right)±\sqrt{31}i}{2}$$

The opposite of -3 is 3.

$$x=\frac{3±\sqrt{31}i}{2}$$

Now solve the equation x=\frac{3±\sqrt{31}i}{2} when ± is plus. Add 3 to i\sqrt{31}.

$$x=\frac{3+\sqrt{31}i}{2}$$

Now solve the equation x=\frac{3±\sqrt{31}i}{2} when ± is minus. Subtract i\sqrt{31} from 3.

$$x=\frac{-\sqrt{31}i+3}{2}$$

The equation is now solved.

$$x=\frac{3+\sqrt{31}i}{2} x=\frac{-\sqrt{31}i+3}{2}$$

Steps for Completing the Square

Variable x cannot be equal to 4 since division by zero is not defined. Multiply both sides of the equation by x-4.

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

Subtract 10 from both sides. Anything subtracted from zero gives its negation.

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

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

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

Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-3x+\frac{9}{4}=-10+\frac{9}{4}$$

Add -10 to \frac{9}{4}.

$$x^{2}-3x+\frac{9}{4}=-\frac{31}{4}$$

Factor x^{2}-3x+\frac{9}{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}.

$$\left(x-\frac{3}{2}\right)^{2}=-\frac{31}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{2}\right)^{2}}=\sqrt{-\frac{31}{4}}$$

Simplify.

$$x-\frac{3}{2}=\frac{\sqrt{31}i}{2} x-\frac{3}{2}=-\frac{\sqrt{31}i}{2}$$

Add \frac{3}{2} to both sides of the equation.

$$x=\frac{3+\sqrt{31}i}{2} x=\frac{-\sqrt{31}i+3}{2}$$