4.9t^2+10t=3.5

Solution for 4.9t^2+10t=3.5 equation:

4.9t^2+10t=3.5

We move all terms to the left:

4.9t^2+10t-(3.5)=0

We add all the numbers together, and all the variables

4.9t^2+10t-3.5=0

a = 4.9; b = 10; c = -3.5;
Δ = b2-4ac
Δ = 102-4·4.9·(-3.5)
Δ = 168.6
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{-(10)-\sqrt{168.6}}{2*4.9}=\frac{-10-\sqrt{168.6}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+\sqrt{168.6}}{2*4.9}=\frac{-10+\sqrt{168.6}}{9.8}
$