Is that supposed to be x² -4x -5 = 0 or x²-4-5=0 as written? We'll do both.
As written it's
x²=9 so x=±3, no quadratic formula required.
Let's assume you meant
x² -4x - 5 = 0
So much temptation. This factors as
(x-5)(x+1)=0 so roots x=5, x=-1. But that didn't use the quadratic formula..
There's a shortcut to the quadratic formula called the Shakespeare Quadratic Formula (2b or -2b) that's useful when the linear term is even.
In that case x²-2bx+c has zeros at z=b±√(b²-c)
x = 2 ± √(4 - -5) = 2 ± √9 = 2 ± 3 = { -1 , 5 }
OK, let's do it like they want
x² - 4x - 5 = 0
We have a=1, b=-4, c=-5 and the quadratic formula
[tex]x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}[/tex]
[tex]x = \dfrac{-(-4) \pm \sqrt{(-4)^2-4(1)(-5)}}{2(1)} = \frac 1 2 (4 \pm \sqrt{36}) = \frac 1 2 (4 \pm 6) = \{-1, 5\}[/tex]
Answer: x=-1 or x=5
