-4.9t^2+30t+13=0

Solution for -4.9t^2+30t+13=0 equation:

-4.9t^2+30t+13=0

a = -4.9; b = 30; c = +13;
Δ = b2-4ac
Δ = 302-4·(-4.9)·13
Δ = 1154.8
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{-(30)-\sqrt{1154.8}}{2*-4.9}=\frac{-30-\sqrt{1154.8}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(30)+\sqrt{1154.8}}{2*-4.9}=\frac{-30+\sqrt{1154.8}}{-9.8}
$