-0.25y^2+16=0

Solution for -0.25y^2+16=0 equation:

-0.25y^2+16=0

a = -0.25; b = 0; c = +16;
Δ = b2-4ac
Δ = 02-4·(-0.25)·16
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4}{2*-0.25}=\frac{-4}{-0.5}
=+8
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4}{2*-0.25}=\frac{4}{-0.5}
=-8
$