4.9t2+98t=205

Solution for 4.9t2+98t=205 equation:

4.9t^2+98t=205

We move all terms to the left:

4.9t^2+98t-(205)=0

a = 4.9; b = 98; c = -205;
Δ = b2-4ac
Δ = 982-4·4.9·(-205)
Δ = 13622
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:

$\sqrt{\Delta}=\sqrt{13622}=\sqrt{49*278}=\sqrt{49}*\sqrt{278}=7\sqrt{278}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(98)-7\sqrt{278}}{2*4.9}=\frac{-98-7\sqrt{278}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(98)+7\sqrt{278}}{2*4.9}=\frac{-98+7\sqrt{278}}{9.8}
$