5r^2+10r-2=0

Solution for 5r^2+10r-2=0 equation:

5r^2+10r-2=0

a = 5; b = 10; c = -2;
Δ = b2-4ac
Δ = 102-4·5·(-2)
Δ = 140
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:

$\sqrt{\Delta}=\sqrt{140}=\sqrt{4*35}=\sqrt{4}*\sqrt{35}=2\sqrt{35}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{35}}{2*5}=\frac{-10-2\sqrt{35}}{10}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{35}}{2*5}=\frac{-10+2\sqrt{35}}{10}
$