-0,3y^2+8=0

Solution for -0,3y^2+8=0 equation:

-0.3y^2+8=0

a = -0.3; b = 0; c = +8;
Δ = b2-4ac
Δ = 02-4·(-0.3)·8
Δ = 9.6
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-\sqrt{9.6}}{2*-0.3}=\frac{0-\sqrt{9.6}}{-0.6}
=-\frac{\sqrt{}}{-0.6}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+\sqrt{9.6}}{2*-0.3}=\frac{0+\sqrt{9.6}}{-0.6}
=\frac{\sqrt{}}{-0.6}
$