0=4.9t^2+5t+-45

Solution for 0=4.9t^2+5t+-45 equation:

0=4.9t^2+5t+-45

We move all terms to the left:

0-(4.9t^2+5t+-45)=0

We add all the numbers together, and all the variables

-(4.9t^2+5t+-45)=0

We use the square of the difference formula

-(4.9t^2+5t-45)=0

We get rid of parentheses

-4.9t^2-5t+45=0

a = -4.9; b = -5; c = +45;
Δ = b2-4ac
Δ = -52-4·(-4.9)·45
Δ = 907
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:

$\sqrt{\Delta}=\sqrt{907}=\sqrt{1*907}=\sqrt{1}*\sqrt{907}=1\sqrt{907}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-1\sqrt{907}}{2*-4.9}=\frac{5-1\sqrt{907}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+1\sqrt{907}}{2*-4.9}=\frac{5+1\sqrt{907}}{-9.8}
$