-4.9t^2+42t+2.5=0

Solution for -4.9t^2+42t+2.5=0 equation:

-4.9t^2+42t+2.5=0

a = -4.9; b = 42; c = +2.5;
Δ = b2-4ac
Δ = 422-4·(-4.9)·2.5
Δ = 1813
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1813}=\sqrt{49*37}=\sqrt{49}*\sqrt{37}=7\sqrt{37}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-7\sqrt{37}}{2*-4.9}=\frac{-42-7\sqrt{37}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+7\sqrt{37}}{2*-4.9}=\frac{-42+7\sqrt{37}}{-9.8}
$