-0,4x^2+40=0

Solution for -0,4x^2+40=0 equation:

-0.4x^2+40=0

a = -0.4; b = 0; c = +40;
Δ = b2-4ac
Δ = 02-4·(-0.4)·40
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8}{2*-0.4}=\frac{-8}{-0.8}
=+10
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8}{2*-0.4}=\frac{8}{-0.8}
=-10
$