4t^2+24t+6=0

Solution for 4t^2+24t+6=0 equation:

4t^2+24t+6=0

a = 4; b = 24; c = +6;
Δ = b2-4ac
Δ = 242-4·4·6
Δ = 480
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:

$\sqrt{\Delta}=\sqrt{480}=\sqrt{16*30}=\sqrt{16}*\sqrt{30}=4\sqrt{30}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-4\sqrt{30}}{2*4}=\frac{-24-4\sqrt{30}}{8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+4\sqrt{30}}{2*4}=\frac{-24+4\sqrt{30}}{8}
$