4t2+48t+3=0

Solution for 4t2+48t+3=0 equation:

4t^2+48t+3=0

a = 4; b = 48; c = +3;
Δ = b2-4ac
Δ = 482-4·4·3
Δ = 2256
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:

$\sqrt{\Delta}=\sqrt{2256}=\sqrt{16*141}=\sqrt{16}*\sqrt{141}=4\sqrt{141}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-4\sqrt{141}}{2*4}=\frac{-48-4\sqrt{141}}{8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+4\sqrt{141}}{2*4}=\frac{-48+4\sqrt{141}}{8}
$