0=4.9t^2+10t-10

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

0=4.9t^2+10t-10

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

-4.9t^2-10t+10=0

a = -4.9; b = -10; c = +10;
Δ = b2-4ac
Δ = -102-4·(-4.9)·10
Δ = 296
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:

$\sqrt{\Delta}=\sqrt{296}=\sqrt{4*74}=\sqrt{4}*\sqrt{74}=2\sqrt{74}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{74}}{2*-4.9}=\frac{10-2\sqrt{74}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{74}}{2*-4.9}=\frac{10+2\sqrt{74}}{-9.8}
$