(k-7)(k-4)=0

Solution for (k-7)(k-4)=0 equation:

(k-7)(k-4)=0

We multiply parentheses ..

(+k^2-4k-7k+28)=0

We get rid of parentheses

k^2-4k-7k+28=0

We add all the numbers together, and all the variables

k^2-11k+28=0

a = 1; b = -11; c = +28;
Δ = b2-4ac
Δ = -112-4·1·28
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-3}{2*1}=\frac{8}{2}
=4
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+3}{2*1}=\frac{14}{2}
=7
$