(7k-4)(k-3)=0

Solution for (7k-4)(k-3)=0 equation:

(7k-4)(k-3)=0

We multiply parentheses ..

(+7k^2-21k-4k+12)=0

We get rid of parentheses

7k^2-21k-4k+12=0

We add all the numbers together, and all the variables

7k^2-25k+12=0

a = 7; b = -25; c = +12;
Δ = b2-4ac
Δ = -252-4·7·12
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-25)-17}{2*7}=\frac{8}{14}
=4/7
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+17}{2*7}=\frac{42}{14}
=3
$