t^2+6t-1=0

Solution for t^2+6t-1=0 equation:

t^2+6t-1=0

a = 1; b = 6; c = -1;
Δ = b2-4ac
Δ = 62-4·1·(-1)
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{-(6)-2\sqrt{10}}{2*1}=\frac{-6-2\sqrt{10}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{10}}{2*1}=\frac{-6+2\sqrt{10}}{2}
$