(k-46)(k-7)=0

Solution for (k-46)(k-7)=0 equation:

(k-46)(k-7)=0

We multiply parentheses ..

(+k^2-7k-46k+322)=0

We get rid of parentheses

k^2-7k-46k+322=0

We add all the numbers together, and all the variables

k^2-53k+322=0

a = 1; b = -53; c = +322;
Δ = b2-4ac
Δ = -532-4·1·322
Δ = 1521
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1521}=39$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-53)-39}{2*1}=\frac{14}{2}
=7
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-53)+39}{2*1}=\frac{92}{2}
=46
$