4.9t^2+6t-150=0

Solution for 4.9t^2+6t-150=0 equation:

4.9t^2+6t-150=0

a = 4.9; b = 6; c = -150;
Δ = b2-4ac
Δ = 62-4·4.9·(-150)
Δ = 2976
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:

$\sqrt{\Delta}=\sqrt{2976}=\sqrt{16*186}=\sqrt{16}*\sqrt{186}=4\sqrt{186}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-4\sqrt{186}}{2*4.9}=\frac{-6-4\sqrt{186}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+4\sqrt{186}}{2*4.9}=\frac{-6+4\sqrt{186}}{9.8}
$