4.9t^2=10t+45

Solution for 4.9t^2=10t+45 equation:

4.9t^2=10t+45

We move all terms to the left:

4.9t^2-(10t+45)=0

We get rid of parentheses

4.9t^2-10t-45=0

a = 4.9; b = -10; c = -45;
Δ = b2-4ac
Δ = -102-4·4.9·(-45)
Δ = 982
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:

$\sqrt{\Delta}=\sqrt{982}=\sqrt{1*982}=\sqrt{1}*\sqrt{982}=1\sqrt{982}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-1\sqrt{982}}{2*4.9}=\frac{10-1\sqrt{982}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+1\sqrt{982}}{2*4.9}=\frac{10+1\sqrt{982}}{9.8}
$