Solve xx-3=7x-20

$$x x - 3 = 7 x - 20$$

Solve for x (complex solution)

$$x=\frac{7+\sqrt{19}i}{2}\approx 3.5+2.179449472i$$

$$x=\frac{-\sqrt{19}i+7}{2}\approx 3.5-2.179449472i$$

Steps Using the Quadratic Formula

Multiply x and x to get x^{2}.

$$x^{2}-3=7x-20$$

Subtract 7x from both sides.

$$x^{2}-3-7x=-20$$

Add 20 to both sides.

$$x^{2}-3-7x+20=0$$

Add -3 and 20 to get 17.

$$x^{2}+17-7x=0$$

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.

$$x^{2}-7x+17=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -7 for b, and 17 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 17}}{2}$$

Square -7.

$$x=\frac{-\left(-7\right)±\sqrt{49-4\times 17}}{2}$$

Multiply -4 times 17.

$$x=\frac{-\left(-7\right)±\sqrt{49-68}}{2}$$

Add 49 to -68.

$$x=\frac{-\left(-7\right)±\sqrt{-19}}{2}$$

Take the square root of -19.

$$x=\frac{-\left(-7\right)±\sqrt{19}i}{2}$$

The opposite of -7 is 7.

$$x=\frac{7±\sqrt{19}i}{2}$$

Now solve the equation x=\frac{7±\sqrt{19}i}{2} when ± is plus. Add 7 to i\sqrt{19}.

$$x=\frac{7+\sqrt{19}i}{2}$$

Now solve the equation x=\frac{7±\sqrt{19}i}{2} when ± is minus. Subtract i\sqrt{19} from 7.

$$x=\frac{-\sqrt{19}i+7}{2}$$

The equation is now solved.

$$x=\frac{7+\sqrt{19}i}{2} x=\frac{-\sqrt{19}i+7}{2}$$

Steps for Completing the Square

Multiply x and x to get x^{2}.

$$x^{2}-3=7x-20$$

Subtract 7x from both sides.

$$x^{2}-3-7x=-20$$

Add 3 to both sides.

$$x^{2}-7x=-20+3$$

Add -20 and 3 to get -17.

$$x^{2}-7x=-17$$

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

$$x^{2}-7x+\left(-\frac{7}{2}\right)^{2}=-17+\left(-\frac{7}{2}\right)^{2}$$

Square -\frac{7}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-7x+\frac{49}{4}=-17+\frac{49}{4}$$

Add -17 to \frac{49}{4}.

$$x^{2}-7x+\frac{49}{4}=-\frac{19}{4}$$

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

$$\left(x-\frac{7}{2}\right)^{2}=-\frac{19}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{2}\right)^{2}}=\sqrt{-\frac{19}{4}}$$

Simplify.

$$x-\frac{7}{2}=\frac{\sqrt{19}i}{2} x-\frac{7}{2}=-\frac{\sqrt{19}i}{2}$$

Add \frac{7}{2} to both sides of the equation.

$$x=\frac{7+\sqrt{19}i}{2} x=\frac{-\sqrt{19}i+7}{2}$$