-0.08(t^2)+4.48=0

Solution for -0.08(t^2)+4.48=0 equation:

-0.08(t^2)+4.48=0

a = -0.08; b = 0; c = +4.48;
Δ = b2-4ac
Δ = 02-4·(-0.08)·4.48
Δ = 1.4336
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-\sqrt{1.4336}}{2*-0.08}=\frac{0-\sqrt{1.4336}}{-0.16}
=-\frac{\sqrt{}}{-0.16}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+\sqrt{1.4336}}{2*-0.08}=\frac{0+\sqrt{1.4336}}{-0.16}
=\frac{\sqrt{}}{-0.16}
$