c^2+6c-5=0

Solution for c^2+6c-5=0 equation:

c^2+6c-5=0

a = 1; b = 6; c = -5;
Δ = b2-4ac
Δ = 62-4·1·(-5)
Δ = 56
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{56}=\sqrt{4*14}=\sqrt{4}*\sqrt{14}=2\sqrt{14}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{14}}{2*1}=\frac{-6-2\sqrt{14}}{2}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{14}}{2*1}=\frac{-6+2\sqrt{14}}{2}
$