(-0.21t^2)+2t+25=0

Solution for (-0.21t^2)+2t+25=0 equation:

(-0.21t^2)+2t+25=0

We get rid of parentheses

-0.21t^2+2t+25=0

a = -0.21; b = 2; c = +25;
Δ = b2-4ac
Δ = 22-4·(-0.21)·25
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-5}{2*-0.21}=\frac{-7}{-0.42}
=16+0.28/0.42
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+5}{2*-0.21}=\frac{3}{-0.42}
=-7+0.06/0.42
$