12+3t-t^2=0

Solution for 12+3t-t^2=0 equation:

12+3t-t^2=0

We add all the numbers together, and all the variables

-1t^2+3t+12=0

a = -1; b = 3; c = +12;
Δ = b2-4ac
Δ = 32-4·(-1)·12
Δ = 57
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{-(3)-\sqrt{57}}{2*-1}=\frac{-3-\sqrt{57}}{-2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{57}}{2*-1}=\frac{-3+\sqrt{57}}{-2}
$