1=-4.9t^2+21t+1.2

Solution for 1=-4.9t^2+21t+1.2 equation:

1=-4.9t^2+21t+1.2

We move all terms to the left:

1-(-4.9t^2+21t+1.2)=0

We get rid of parentheses

4.9t^2-21t-1.2+1=0

We add all the numbers together, and all the variables

4.9t^2-21t-0.2=0

a = 4.9; b = -21; c = -0.2;
Δ = b2-4ac
Δ = -212-4·4.9·(-0.2)
Δ = 444.92
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{-(-21)-\sqrt{444.92}}{2*4.9}=\frac{21-\sqrt{444.92}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+\sqrt{444.92}}{2*4.9}=\frac{21+\sqrt{444.92}}{9.8}
$