20k+32=2k2

Solution for 20k+32=2k2 equation:

20k+32=2k^2

We move all terms to the left:

20k+32-(2k^2)=0

determiningTheFunctionDomain
-2k^2+20k+32=0

a = -2; b = 20; c = +32;
Δ = b2-4ac
Δ = 202-4·(-2)·32
Δ = 656
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:

$\sqrt{\Delta}=\sqrt{656}=\sqrt{16*41}=\sqrt{16}*\sqrt{41}=4\sqrt{41}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{41}}{2*-2}=\frac{-20-4\sqrt{41}}{-4}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{41}}{2*-2}=\frac{-20+4\sqrt{41}}{-4}
$