4.9t^2+8t-57=0

Solution for 4.9t^2+8t-57=0 equation:

4.9t^2+8t-57=0

a = 4.9; b = 8; c = -57;
Δ = b2-4ac
Δ = 82-4·4.9·(-57)
Δ = 1181.2
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{-(8)-\sqrt{1181.2}}{2*4.9}=\frac{-8-\sqrt{1181.2}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+\sqrt{1181.2}}{2*4.9}=\frac{-8+\sqrt{1181.2}}{9.8}
$