t=4.9t^2+1.25t-12-0

Solution for t=4.9t^2+1.25t-12-0 equation:

t=4.9t^2+1.25t-12-0

We move all terms to the left:

t-(4.9t^2+1.25t-12-0)=0

We get rid of parentheses

-4.9t^2+t-1.25t+12+0=0

We add all the numbers together, and all the variables

-4.9t^2-0.25t+12=0

a = -4.9; b = -0.25; c = +12;
Δ = b2-4ac
Δ = -0.252-4·(-4.9)·12
Δ = 235.2625
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{-(-0.25)-\sqrt{235.2625}}{2*-4.9}=\frac{0.25-\sqrt{235.2625}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-0.25)+\sqrt{235.2625}}{2*-4.9}=\frac{0.25+\sqrt{235.2625}}{-9.8}
$