-0,0023x^2+40=0

Solution for -0,0023x^2+40=0 equation:

-0.0023x^2+40=0

a = -0.0023; b = 0; c = +40;
Δ = b2-4ac
Δ = 02-4·(-0.0023)·40
Δ = 0.368
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)-\sqrt{0.368}}{2*-0.0023}=\frac{0-\sqrt{0.368}}{-0.0046}
=-\frac{\sqrt{}}{-0.0046}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+\sqrt{0.368}}{2*-0.0023}=\frac{0+\sqrt{0.368}}{-0.0046}
=\frac{\sqrt{}}{-0.0046}
$