0.35x^2-32x+500=0

Solution for 0.35x^2-32x+500=0 equation:

0.35x^2-32x+500=0

a = 0.35; b = -32; c = +500;
Δ = b2-4ac
Δ = -322-4·0.35·500
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-18}{2*0.35}=\frac{14}{0.7}
=20
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+18}{2*0.35}=\frac{50}{0.7}
=71+0.3/0.7
$