7k^2+20k=3

Solution for 7k^2+20k=3 equation:

7k^2+20k=3

We move all terms to the left:

7k^2+20k-(3)=0

a = 7; b = 20; c = -3;
Δ = b2-4ac
Δ = 202-4·7·(-3)
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-22}{2*7}=\frac{-42}{14}
=-3
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+22}{2*7}=\frac{2}{14}
=1/7
$