2t+4,9t^2-10,5=0

Solution for 2t+4,9t^2-10,5=0 equation:

2t+4.9t^2-10.5=0

a = 4.9; b = 2; c = -10.5;
Δ = b2-4ac
Δ = 22-4·4.9·(-10.5)
Δ = 209.8
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{-(2)-\sqrt{209.8}}{2*4.9}=\frac{-2-\sqrt{209.8}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+\sqrt{209.8}}{2*4.9}=\frac{-2+\sqrt{209.8}}{9.8}
$