-4t+0,48t^2-25=0

Solution for -4t+0,48t^2-25=0 equation:

-4t+0.48t^2-25=0

a = 0.48; b = -4; c = -25;
Δ = b2-4ac
Δ = -42-4·0.48·(-25)
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-8}{2*0.48}=\frac{-4}{0.96}
=-4+0.16/0.96
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+8}{2*0.48}=\frac{12}{0.96}
=12+0.48/0.96
$