2n^2+10n-2=0

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

2n^2+10n-2=0

a = 2; b = 10; c = -2;
Δ = b2-4ac
Δ = 102-4·2·(-2)
Δ = 116
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{116}=\sqrt{4*29}=\sqrt{4}*\sqrt{29}=2\sqrt{29}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{29}}{2*2}=\frac{-10-2\sqrt{29}}{4}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{29}}{2*2}=\frac{-10+2\sqrt{29}}{4}
$