4t^2+12t+1=0

Solution for 4t^2+12t+1=0 equation:

4t^2+12t+1=0

a = 4; b = 12; c = +1;
Δ = b2-4ac
Δ = 122-4·4·1
Δ = 128
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:

$\sqrt{\Delta}=\sqrt{128}=\sqrt{64*2}=\sqrt{64}*\sqrt{2}=8\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-8\sqrt{2}}{2*4}=\frac{-12-8\sqrt{2}}{8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+8\sqrt{2}}{2*4}=\frac{-12+8\sqrt{2}}{8}
$