-0.05x^2+x+40=0

Solution for -0.05x^2+x+40=0 equation:

-0.05x^2+x+40=0

a = -0.05; b = 1; c = +40;
Δ = b2-4ac
Δ = 12-4·(-0.05)·40
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-3}{2*-0.05}=\frac{-4}{-0.1}
=+40
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+3}{2*-0.05}=\frac{2}{-0.1}
=-20
$