k(7k+2)=5

Solution for k(7k+2)=5 equation:

k(7k+2)=5

We move all terms to the left:

k(7k+2)-(5)=0

We multiply parentheses

7k^2+2k-5=0

a = 7; b = 2; c = -5;
Δ = b2-4ac
Δ = 22-4·7·(-5)
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-12}{2*7}=\frac{-14}{14}
=-1
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+12}{2*7}=\frac{10}{14}
=5/7
$