4(6t+1)t=3=

Solution for 4(6t+1)t=3= equation:

4(6t+1)t=3=

We move all terms to the left:

4(6t+1)t-(3)=0

We multiply parentheses

24t^2+4t-3=0

a = 24; b = 4; c = -3;
Δ = b2-4ac
Δ = 42-4·24·(-3)
Δ = 304
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{304}=\sqrt{16*19}=\sqrt{16}*\sqrt{19}=4\sqrt{19}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{19}}{2*24}=\frac{-4-4\sqrt{19}}{48}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{19}}{2*24}=\frac{-4+4\sqrt{19}}{48}
$