2k^2+20k=3

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

2k^2+20k=3

We move all terms to the left:

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

a = 2; b = 20; c = -3;
Δ = b2-4ac
Δ = 202-4·2·(-3)
Δ = 424
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{424}=\sqrt{4*106}=\sqrt{4}*\sqrt{106}=2\sqrt{106}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-2\sqrt{106}}{2*2}=\frac{-20-2\sqrt{106}}{4}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+2\sqrt{106}}{2*2}=\frac{-20+2\sqrt{106}}{4}
$