2a^2-65a+32=0

Solution for 2a^2-65a+32=0 equation:

2a^2-65a+32=0

a = 2; b = -65; c = +32;
Δ = b2-4ac
Δ = -652-4·2·32
Δ = 3969
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3969}=63$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-65)-63}{2*2}=\frac{2}{4}
=1/2
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-65)+63}{2*2}=\frac{128}{4}
=32
$