2n^2+24n+10=0

Solution for 2n^2+24n+10=0 equation:

2n^2+24n+10=0

a = 2; b = 24; c = +10;
Δ = b2-4ac
Δ = 242-4·2·10
Δ = 496
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:

$\sqrt{\Delta}=\sqrt{496}=\sqrt{16*31}=\sqrt{16}*\sqrt{31}=4\sqrt{31}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-4\sqrt{31}}{2*2}=\frac{-24-4\sqrt{31}}{4}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+4\sqrt{31}}{2*2}=\frac{-24+4\sqrt{31}}{4}
$