-4.9t^2-8t+90=0

Solution for -4.9t^2-8t+90=0 equation:

-4.9t^2-8t+90=0

a = -4.9; b = -8; c = +90;
Δ = b2-4ac
Δ = -82-4·(-4.9)·90
Δ = 1828
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:

$\sqrt{\Delta}=\sqrt{1828}=\sqrt{4*457}=\sqrt{4}*\sqrt{457}=2\sqrt{457}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-2\sqrt{457}}{2*-4.9}=\frac{8-2\sqrt{457}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+2\sqrt{457}}{2*-4.9}=\frac{8+2\sqrt{457}}{-9.8}
$