2r^2+6r-2=0

Solution for 2r^2+6r-2=0 equation:

2r^2+6r-2=0

a = 2; b = 6; c = -2;
Δ = b2-4ac
Δ = 62-4·2·(-2)
Δ = 52
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:

$\sqrt{\Delta}=\sqrt{52}=\sqrt{4*13}=\sqrt{4}*\sqrt{13}=2\sqrt{13}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{13}}{2*2}=\frac{-6-2\sqrt{13}}{4}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{13}}{2*2}=\frac{-6+2\sqrt{13}}{4}
$