k(7k-3)=0

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

k(7k-3)=0

We multiply parentheses

7k^2-3k=0

a = 7; b = -3; c = 0;
Δ = b2-4ac
Δ = -32-4·7·0
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{-(-3)-3}{2*7}=\frac{0}{14}
=0
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3}{2*7}=\frac{6}{14}
=3/7
$