4.9t^2+9t=650

Solution for 4.9t^2+9t=650 equation:

4.9t^2+9t=650

We move all terms to the left:

4.9t^2+9t-(650)=0

a = 4.9; b = 9; c = -650;
Δ = b2-4ac
Δ = 92-4·4.9·(-650)
Δ = 12821
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:

$\sqrt{\Delta}=\sqrt{12821}=\sqrt{1*12821}=\sqrt{1}*\sqrt{12821}=1\sqrt{12821}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-1\sqrt{12821}}{2*4.9}=\frac{-9-1\sqrt{12821}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+1\sqrt{12821}}{2*4.9}=\frac{-9+1\sqrt{12821}}{9.8}
$