Solve (x+3)(x-5)=4(x-5)

$$(x+3)(x-5)=4(x-5)$$

Solve for x

$$x=1$$

$$x=5$$

Steps Using the Quadratic Formula

Use the distributive property to multiply x+3 by x-5 and combine like terms.

$$x^{2}-2x-15=4\left(x-5\right)$$

Use the distributive property to multiply 4 by x-5.

$$x^{2}-2x-15=4x-20$$

Subtract 4x from both sides.

$$x^{2}-2x-15-4x=-20$$

Combine -2x and -4x to get -6x.

$$x^{2}-6x-15=-20$$

Add 20 to both sides.

$$x^{2}-6x-15+20=0$$

Add -15 and 20 to get 5.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -6 for b, and 5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 5}}{2}$$

Square -6.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\times 5}}{2}$$

Multiply -4 times 5.

$$x=\frac{-\left(-6\right)±\sqrt{36-20}}{2}$$

Add 36 to -20.

$$x=\frac{-\left(-6\right)±\sqrt{16}}{2}$$

Take the square root of 16.

$$x=\frac{-\left(-6\right)±4}{2}$$

The opposite of -6 is 6.

$$x=\frac{6±4}{2}$$

Now solve the equation x=\frac{6±4}{2} when ± is plus. Add 6 to 4.

$$x=\frac{10}{2}$$

Divide 10 by 2.

$$x=5$$

Now solve the equation x=\frac{6±4}{2} when ± is minus. Subtract 4 from 6.

$$x=\frac{2}{2}$$

Divide 2 by 2.

$$x=1$$

The equation is now solved.

$$x=5 x=1$$

Steps for Completing the Square

Use the distributive property to multiply x+3 by x-5 and combine like terms.

$$x^{2}-2x-15=4\left(x-5\right)$$

Use the distributive property to multiply 4 by x-5.

$$x^{2}-2x-15=4x-20$$

Subtract 4x from both sides.

$$x^{2}-2x-15-4x=-20$$

Combine -2x and -4x to get -6x.

$$x^{2}-6x-15=-20$$

Add 15 to both sides.

$$x^{2}-6x=-20+15$$

Add -20 and 15 to get -5.

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

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.

$$x^{2}-6x+\left(-3\right)^{2}=-5+\left(-3\right)^{2}$$

Square -3.

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

Add -5 to 9.

$$x^{2}-6x+9=4$$

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-3=2 x-3=-2$$

Add 3 to both sides of the equation.

$$x=5 x=1$$