5t^2+126t+26=0

Solution for 5t^2+126t+26=0 equation:

5t^2+126t+26=0

a = 5; b = 126; c = +26;
Δ = b2-4ac
Δ = 1262-4·5·26
Δ = 15356
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:

$\sqrt{\Delta}=\sqrt{15356}=\sqrt{4*3839}=\sqrt{4}*\sqrt{3839}=2\sqrt{3839}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(126)-2\sqrt{3839}}{2*5}=\frac{-126-2\sqrt{3839}}{10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(126)+2\sqrt{3839}}{2*5}=\frac{-126+2\sqrt{3839}}{10}
$