7t2+6t+1=0

Solution for 7t2+6t+1=0 equation:

7t^2+6t+1=0

a = 7; b = 6; c = +1;
Δ = b2-4ac
Δ = 62-4·7·1
Δ = 8
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:

$\sqrt{\Delta}=\sqrt{8}=\sqrt{4*2}=\sqrt{4}*\sqrt{2}=2\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{2}}{2*7}=\frac{-6-2\sqrt{2}}{14}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{2}}{2*7}=\frac{-6+2\sqrt{2}}{14}
$