Solve (x+5)(x-5)-5(x-6)=x-9

$$(x+5)(x-5)-5(x-6)=x-9$$

Solve for x (complex solution)

$$x=3+\sqrt{5}i\approx 3+2.236067977i$$

$$x=-\sqrt{5}i+3\approx 3-2.236067977i$$

Steps Using the Quadratic Formula

Consider \left(x+5\right)\left(x-5\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 5.

$$x^{2}-25-5\left(x-6\right)=x-9$$

Use the distributive property to multiply -5 by x-6.

$$x^{2}-25-5x+30=x-9$$

Add -25 and 30 to get 5.

$$x^{2}+5-5x=x-9$$

Subtract x from both sides.

$$x^{2}+5-5x-x=-9$$

Combine -5x and -x to get -6x.

$$x^{2}+5-6x=-9$$

Add 9 to both sides.

$$x^{2}+5-6x+9=0$$

Add 5 and 9 to get 14.

$$x^{2}+14-6x=0$$

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.

$$x^{2}-6x+14=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -6 for b, and 14 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 14}}{2}$$

Square -6.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\times 14}}{2}$$

Multiply -4 times 14.

$$x=\frac{-\left(-6\right)±\sqrt{36-56}}{2}$$

Add 36 to -56.

$$x=\frac{-\left(-6\right)±\sqrt{-20}}{2}$$

Take the square root of -20.

$$x=\frac{-\left(-6\right)±2\sqrt{5}i}{2}$$

The opposite of -6 is 6.

$$x=\frac{6±2\sqrt{5}i}{2}$$

Now solve the equation x=\frac{6±2\sqrt{5}i}{2} when ± is plus. Add 6 to 2i\sqrt{5}.

$$x=\frac{6+2\sqrt{5}i}{2}$$

Divide 6+2i\sqrt{5} by 2.

$$x=3+\sqrt{5}i$$

Now solve the equation x=\frac{6±2\sqrt{5}i}{2} when ± is minus. Subtract 2i\sqrt{5} from 6.

$$x=\frac{-2\sqrt{5}i+6}{2}$$

Divide 6-2i\sqrt{5} by 2.

$$x=-\sqrt{5}i+3$$

The equation is now solved.

$$x=3+\sqrt{5}i x=-\sqrt{5}i+3$$

Steps for Completing the Square

Consider \left(x+5\right)\left(x-5\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 5.

$$x^{2}-25-5\left(x-6\right)=x-9$$

Use the distributive property to multiply -5 by x-6.

$$x^{2}-25-5x+30=x-9$$

Add -25 and 30 to get 5.

$$x^{2}+5-5x=x-9$$

Subtract x from both sides.

$$x^{2}+5-5x-x=-9$$

Combine -5x and -x to get -6x.

$$x^{2}+5-6x=-9$$

Subtract 5 from both sides.

$$x^{2}-6x=-9-5$$

Subtract 5 from -9 to get -14.

$$x^{2}-6x=-14$$

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

$$x^{2}-6x+\left(-3\right)^{2}=-14+\left(-3\right)^{2}$$

Square -3.

$$x^{2}-6x+9=-14+9$$

Add -14 to 9.

$$x^{2}-6x+9=-5$$

Factor x^{2}-6x+9. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-3\right)^{2}=-5$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-3\right)^{2}}=\sqrt{-5}$$

Simplify.

$$x-3=\sqrt{5}i x-3=-\sqrt{5}i$$

Add 3 to both sides of the equation.

$$x=3+\sqrt{5}i x=-\sqrt{5}i+3$$