32n^2=98

Solution for 32n^2=98 equation:

32n^2=98

We move all terms to the left:

32n^2-(98)=0

a = 32; b = 0; c = -98;
Δ = b2-4ac
Δ = 02-4·32·(-98)
Δ = 12544
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{12544}=112$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-112}{2*32}=\frac{-112}{64}
=-1+3/4
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+112}{2*32}=\frac{112}{64}
=1+3/4
$