12t^2-148t+48=0

Solution for 12t^2-148t+48=0 equation:

12t^2-148t+48=0

a = 12; b = -148; c = +48;
Δ = b2-4ac
Δ = -1482-4·12·48
Δ = 19600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{19600}=140$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-148)-140}{2*12}=\frac{8}{24}
=1/3
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-148)+140}{2*12}=\frac{288}{24}
=12
$