r^2+10r-2=0

Solution for r^2+10r-2=0 equation:

r^2+10r-2=0

a = 1; b = 10; c = -2;
Δ = b2-4ac
Δ = 102-4·1·(-2)
Δ = 108
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{108}=\sqrt{36*3}=\sqrt{36}*\sqrt{3}=6\sqrt{3}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-6\sqrt{3}}{2*1}=\frac{-10-6\sqrt{3}}{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+6\sqrt{3}}{2*1}=\frac{-10+6\sqrt{3}}{2}
$