-0.01x^2+0.5x+6.2=0

Solution for -0.01x^2+0.5x+6.2=0 equation:

-0.01x^2+0.5x+6.2=0

a = -0.01; b = 0.5; c = +6.2;
Δ = b2-4ac
Δ = 0.52-4·(-0.01)·6.2
Δ = 0.498
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0.5)-\sqrt{0.498}}{2*-0.01}=\frac{-0.5-\sqrt{0.498}}{-0.02}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.5)+\sqrt{0.498}}{2*-0.01}=\frac{-0.5+\sqrt{0.498}}{-0.02}
$