-4.9t^2+30t+100=0

Solution for -4.9t^2+30t+100=0 equation:

-4.9t^2+30t+100=0

a = -4.9; b = 30; c = +100;
Δ = b2-4ac
Δ = 302-4·(-4.9)·100
Δ = 2860
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:

$\sqrt{\Delta}=\sqrt{2860}=\sqrt{4*715}=\sqrt{4}*\sqrt{715}=2\sqrt{715}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(30)-2\sqrt{715}}{2*-4.9}=\frac{-30-2\sqrt{715}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(30)+2\sqrt{715}}{2*-4.9}=\frac{-30+2\sqrt{715}}{-9.8}
$