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

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

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

We multiply parentheses ..

(+3k^2-9k-7k+21)=0

We get rid of parentheses

3k^2-9k-7k+21=0

We add all the numbers together, and all the variables

3k^2-16k+21=0

a = 3; b = -16; c = +21;
Δ = b2-4ac
Δ = -162-4·3·21
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-2}{2*3}=\frac{14}{6}
=2+1/3
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+2}{2*3}=\frac{18}{6}
=3
$