4.9t^2=4.9t+12

Solution for 4.9t^2=4.9t+12 equation:

4.9t^2=4.9t+12

We move all terms to the left:

4.9t^2-(4.9t+12)=0

We get rid of parentheses

4.9t^2-4.9t-12=0

a = 4.9; b = -4.9; c = -12;
Δ = b2-4ac
Δ = -4.92-4·4.9·(-12)
Δ = 259.21
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{-(-4.9)-\sqrt{259.21}}{2*4.9}=\frac{4.9-\sqrt{259.21}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4.9)+\sqrt{259.21}}{2*4.9}=\frac{4.9+\sqrt{259.21}}{9.8}
$