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

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

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

We move all terms to the left:

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

4k^2+k-5=0

a = 4; b = 1; c = -5;
Δ = b2-4ac
Δ = 12-4·4·(-5)
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-9}{2*4}=\frac{-10}{8}
=-1+1/4
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+9}{2*4}=\frac{8}{8}
=1
$