a+b=-4 ab=-5

To solve the equation, factor x^{2}-4x-5 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

a=-5 b=1

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. The only such pair is the system solution.

\left(x-5\right)\left(x+1\right)

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

x=5 x=-1

To find equation solutions, solve x-5=0 and x+1=0.

a+b=-4 ab=1\left(-5\right)=-5

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-5. To find a and b, set up a system to be solved.

a=-5 b=1

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. The only such pair is the system solution.

\left(x^{2}-5x\right)+\left(x-5\right)

Rewrite x^{2}-4x-5 as \left(x^{2}-5x\right)+\left(x-5\right).

x\left(x-5\right)+x-5

Factor out x in x^{2}-5x.

\left(x-5\right)\left(x+1\right)

Factor out common term x-5 by using distributive property.

x=5 x=-1

To find equation solutions, solve x-5=0 and x+1=0.

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

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

x=\frac{-\left(-4\right)±\sqrt{16-4\left(-5\right)}}{2}

Square -4.

x=\frac{-\left(-4\right)±\sqrt{16+20}}{2}

Multiply -4 times -5.

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

Add 16 to 20.

x=\frac{-\left(-4\right)±6}{2}

Take the square root of 36.

x=\frac{4±6}{2}

The opposite of -4 is 4.

x=\frac{10}{2}

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

x=5

Divide 10 by 2.

x=-\frac{2}{2}

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

x=-1

Divide -2 by 2.

x=5 x=-1

The equation is now solved.

x^{2}-4x-5=0

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

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

Add 5 to both sides of the equation.

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

Subtracting -5 from itself leaves 0.

x^{2}-4x=5

Subtract -5 from 0.

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

Divide -4, the coefficient of the x term, by 2 to get -2. Then add the square of -2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

x^{2}-4x+4=5+4

Square -2.

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

Add 5 to 4.

\left(x-2\right)^{2}=9

Factor x^{2}-4x+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}.

\sqrt{\left(x-2\right)^{2}}=\sqrt{9}

Take the square root of both sides of the equation.

x-2=3 x-2=-3

Simplify.

x=5 x=-1

Add 2 to both sides of the equation.