2n2+35=435

Solution for 2n2+35=435 equation:

2n^2+35=435

We move all terms to the left:

2n^2+35-(435)=0

We add all the numbers together, and all the variables

2n^2-400=0

a = 2; b = 0; c = -400;
Δ = b2-4ac
Δ = 02-4·2·(-400)
Δ = 3200
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:

$\sqrt{\Delta}=\sqrt{3200}=\sqrt{1600*2}=\sqrt{1600}*\sqrt{2}=40\sqrt{2}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-40\sqrt{2}}{2*2}=\frac{0-40\sqrt{2}}{4}
=-\frac{40\sqrt{2}}{4}
=-10\sqrt{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+40\sqrt{2}}{2*2}=\frac{0+40\sqrt{2}}{4}
=\frac{40\sqrt{2}}{4}
=10\sqrt{2}
$