2n^2+16n-2.5=0

Solution for 2n^2+16n-2.5=0 equation:

2n^2+16n-2.5=0

a = 2; b = 16; c = -2.5;
Δ = b2-4ac
Δ = 162-4·2·(-2.5)
Δ = 276
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:

$\sqrt{\Delta}=\sqrt{276}=\sqrt{4*69}=\sqrt{4}*\sqrt{69}=2\sqrt{69}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-2\sqrt{69}}{2*2}=\frac{-16-2\sqrt{69}}{4}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+2\sqrt{69}}{2*2}=\frac{-16+2\sqrt{69}}{4}
$