(k+1)(4k+5)=0

Solution for (k+1)(4k+5)=0 equation:

(k+1)(4k+5)=0

We multiply parentheses ..

(+4k^2+5k+4k+5)=0

We get rid of parentheses

4k^2+5k+4k+5=0

We add all the numbers together, and all the variables

4k^2+9k+5=0

a = 4; b = 9; c = +5;
Δ = b2-4ac
Δ = 92-4·4·5
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-1}{2*4}=\frac{-10}{8}
=-1+1/4
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+1}{2*4}=\frac{-8}{8}
=-1
$