0=(-4.9t^2)+10t+25

Solution for 0=(-4.9t^2)+10t+25 equation:

0=(-4.9t^2)+10t+25

We move all terms to the left:

0-((-4.9t^2)+10t+25)=0

We add all the numbers together, and all the variables

-((-4.9t^2)+10t+25)=0


We calculate terms in parentheses: -((-4.9t^2)+10t+25), so:

(-4.9t^2)+10t+25

We get rid of parentheses

-4.9t^2+10t+25

Back to the equation:

-(-4.9t^2+10t+25)


We get rid of parentheses

4.9t^2-10t-25=0

a = 4.9; b = -10; c = -25;
Δ = b2-4ac
Δ = -102-4·4.9·(-25)
Δ = 590
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{590}}{2*4.9}=\frac{10-\sqrt{590}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+\sqrt{590}}{2*4.9}=\frac{10+\sqrt{590}}{9.8}
$