t^2+8t+6=0

Solution for t^2+8t+6=0 equation:

t^2+8t+6=0

a = 1; b = 8; c = +6;
Δ = b2-4ac
Δ = 82-4·1·6
Δ = 40
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:

$\sqrt{\Delta}=\sqrt{40}=\sqrt{4*10}=\sqrt{4}*\sqrt{10}=2\sqrt{10}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-2\sqrt{10}}{2*1}=\frac{-8-2\sqrt{10}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2\sqrt{10}}{2*1}=\frac{-8+2\sqrt{10}}{2}
$