Solve x^2-4x-5=0/frac{frac{frac{frac{frac{frac{frac{frac{frac{frac{frac{frac{frac{frac{frac{{}}{}}{}}{}}{}}{}}{}}{}}{}}{}}{}}{}}{}}{}}{}}

$${ x }^{ 2 } -4x-5 = \frac{ 0 }{ \frac{ \frac{ \frac{ \frac{ \frac{ \frac{ \frac{ \frac{ \frac{ \frac{ \frac{ \frac{ \frac{ \frac{ \frac{ }{ } }{ } }{ } }{ } }{ } }{ } }{ } }{ } }{ } }{ } }{ } }{ } }{ } }{ } }{ } }$$

Solve for x

$$x=-1$$

$$x=5$$

Steps Using Factoring

Divide 0 by \frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{1}{1}}{1}}{1}}{1}}{1}}{1}}{1}}{1}}{1}}{1}}{1}}{1}}{1}}{1}}{1} by multiplying 0 by the reciprocal of \frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{1}{1}}{1}}{1}}{1}}{1}}{1}}{1}}{1}}{1}}{1}}{1}}{1}}{1}}{1}}{1}.

$$x^{2}-4x-5=\frac{0}{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{1}{1}}{1}}{1}}{1}}{1}}{1}}{1}}{1}}{1}}{1}}{1}}{1}}{1}}{1}}$$

Anything divided by one gives itself.

$$x^{2}-4x-5=\frac{0}{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{1}{1}}{1}}{1}}{1}}{1}}{1}}{1}}{1}}{1}}{1}}{1}}{1}}{1}}$$

Anything divided by one gives itself.

$$x^{2}-4x-5=\frac{0}{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{1}{1}}{1}}{1}}{1}}{1}}{1}}{1}}{1}}{1}}{1}}{1}}{1}}$$

Anything divided by one gives itself.

$$x^{2}-4x-5=\frac{0}{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{1}{1}}{1}}{1}}{1}}{1}}{1}}{1}}{1}}{1}}{1}}{1}}$$

Anything divided by one gives itself.

$$x^{2}-4x-5=\frac{0}{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{1}{1}}{1}}{1}}{1}}{1}}{1}}{1}}{1}}{1}}{1}}$$

Anything divided by one gives itself.

$$x^{2}-4x-5=\frac{0}{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{1}{1}}{1}}{1}}{1}}{1}}{1}}{1}}{1}}{1}}$$

Anything divided by one gives itself.

$$x^{2}-4x-5=\frac{0}{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{1}{1}}{1}}{1}}{1}}{1}}{1}}{1}}{1}}$$

Anything divided by one gives itself.

$$x^{2}-4x-5=\frac{0}{\frac{\frac{\frac{\frac{\frac{\frac{\frac{1}{1}}{1}}{1}}{1}}{1}}{1}}{1}}$$

Anything divided by one gives itself.

$$x^{2}-4x-5=\frac{0}{\frac{\frac{\frac{\frac{\frac{\frac{1}{1}}{1}}{1}}{1}}{1}}{1}}$$

Anything divided by one gives itself.

$$x^{2}-4x-5=\frac{0}{\frac{\frac{\frac{\frac{\frac{1}{1}}{1}}{1}}{1}}{1}}$$

Anything divided by one gives itself.

$$x^{2}-4x-5=\frac{0}{\frac{\frac{\frac{\frac{1}{1}}{1}}{1}}{1}}$$

Anything divided by one gives itself.

$$x^{2}-4x-5=\frac{0}{\frac{\frac{\frac{1}{1}}{1}}{1}}$$

Anything divided by one gives itself.

$$x^{2}-4x-5=\frac{0}{\frac{\frac{1}{1}}{1}}$$

Anything divided by one gives itself.

$$x^{2}-4x-5=\frac{0}{\frac{1}{1}}$$

Anything divided by one gives itself.

$$x^{2}-4x-5=\frac{0}{1}$$

Anything divided by one gives itself.

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

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.

$$a+b=-4 ab=-5$$

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.

$$a=-5 b=1$$

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

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

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

$$x=5 x=-1$$

Steps Using Factoring By Grouping

Divide 0 by \frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{1}{1}}{1}}{1}}{1}}{1}}{1}}{1}}{1}}{1}}{1}}{1}}{1}}{1}}{1}}{1} by multiplying 0 by the reciprocal of \frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{1}{1}}{1}}{1}}{1}}{1}}{1}}{1}}{1}}{1}}{1}}{1}}{1}}{1}}{1}}{1}.

$$x^{2}-4x-5=\frac{0}{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{1}{1}}{1}}{1}}{1}}{1}}{1}}{1}}{1}}{1}}{1}}{1}}{1}}{1}}{1}}$$

Anything divided by one gives itself.

$$x^{2}-4x-5=\frac{0}{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{1}{1}}{1}}{1}}{1}}{1}}{1}}{1}}{1}}{1}}{1}}{1}}{1}}{1}}$$

Anything divided by one gives itself.

$$x^{2}-4x-5=\frac{0}{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{1}{1}}{1}}{1}}{1}}{1}}{1}}{1}}{1}}{1}}{1}}{1}}{1}}$$

Anything divided by one gives itself.

$$x^{2}-4x-5=\frac{0}{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{1}{1}}{1}}{1}}{1}}{1}}{1}}{1}}{1}}{1}}{1}}{1}}$$

Anything divided by one gives itself.

$$x^{2}-4x-5=\frac{0}{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{1}{1}}{1}}{1}}{1}}{1}}{1}}{1}}{1}}{1}}{1}}$$

Anything divided by one gives itself.

$$x^{2}-4x-5=\frac{0}{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{1}{1}}{1}}{1}}{1}}{1}}{1}}{1}}{1}}{1}}$$

Anything divided by one gives itself.

$$x^{2}-4x-5=\frac{0}{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{1}{1}}{1}}{1}}{1}}{1}}{1}}{1}}{1}}$$

Anything divided by one gives itself.

$$x^{2}-4x-5=\frac{0}{\frac{\frac{\frac{\frac{\frac{\frac{\frac{1}{1}}{1}}{1}}{1}}{1}}{1}}{1}}$$

Anything divided by one gives itself.

$$x^{2}-4x-5=\frac{0}{\frac{\frac{\frac{\frac{\frac{\frac{1}{1}}{1}}{1}}{1}}{1}}{1}}$$

Anything divided by one gives itself.

$$x^{2}-4x-5=\frac{0}{\frac{\frac{\frac{\frac{\frac{1}{1}}{1}}{1}}{1}}{1}}$$

Anything divided by one gives itself.

$$x^{2}-4x-5=\frac{0}{\frac{\frac{\frac{\frac{1}{1}}{1}}{1}}{1}}$$

Anything divided by one gives itself.

$$x^{2}-4x-5=\frac{0}{\frac{\frac{\frac{1}{1}}{1}}{1}}$$

Anything divided by one gives itself.

$$x^{2}-4x-5=\frac{0}{\frac{\frac{1}{1}}{1}}$$

Anything divided by one gives itself.

$$x^{2}-4x-5=\frac{0}{\frac{1}{1}}$$

Anything divided by one gives itself.

$$x^{2}-4x-5=\frac{0}{1}$$

Anything divided by one gives itself.

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

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.

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

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.

$$a=-5 b=1$$

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

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

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

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

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

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

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

$$x=5 x=-1$$

Steps Using the Quadratic Formula

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.

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

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}.

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

Square -4.

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

Multiply -4 times -5.

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

Add 16 to 20.

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

Take the square root of 36.

$$x=\frac{-\left(-4\right)±6}{2}$$

The opposite of -4 is 4.

$$x=\frac{4±6}{2}$$

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

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

Divide 10 by 2.

$$x=5$$

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

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

Divide -2 by 2.

$$x=-1$$

The equation is now solved.

$$x=5 x=-1$$

Steps for Completing the Square

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.

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

Add 5 to both sides of the equation.

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

Subtracting -5 from itself leaves 0.

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

Subtract -5 from 0.

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

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.

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

Square -2.

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

Add 5 to 4.

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

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Add 2 to both sides of the equation.

$$x=5 x=-1$$