7k^2+23k+6=0

Solution for 7k^2+23k+6=0 equation:

7k^2+23k+6=0

a = 7; b = 23; c = +6;
Δ = b2-4ac
Δ = 232-4·7·6
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-19}{2*7}=\frac{-42}{14}
=-3
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+19}{2*7}=\frac{-4}{14}
=-2/7
$