-0.3x^2-4x+40=0

Solution for -0.3x^2-4x+40=0 equation:

-0.3x^2-4x+40=0

a = -0.3; b = -4; c = +40;
Δ = b2-4ac
Δ = -42-4·(-0.3)·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{-(-4)-8}{2*-0.3}=\frac{-4}{-0.6}
=6+0.4/0.6
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+8}{2*-0.3}=\frac{12}{-0.6}
=-20
$