n^2+2n-3.5=0

Solution for n^2+2n-3.5=0 equation:

n^2+2n-3.5=0

a = 1; b = 2; c = -3.5;
Δ = b2-4ac
Δ = 22-4·1·(-3.5)
Δ = 18
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:

$\sqrt{\Delta}=\sqrt{18}=\sqrt{9*2}=\sqrt{9}*\sqrt{2}=3\sqrt{2}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-3\sqrt{2}}{2*1}=\frac{-2-3\sqrt{2}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+3\sqrt{2}}{2*1}=\frac{-2+3\sqrt{2}}{2}
$