28x^2+1=396

Solution for 28x^2+1=396 equation:

28x^2+1=396

We move all terms to the left:

28x^2+1-(396)=0

We add all the numbers together, and all the variables

28x^2-395=0

a = 28; b = 0; c = -395;
Δ = b2-4ac
Δ = 02-4·28·(-395)
Δ = 44240
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{44240}=\sqrt{16*2765}=\sqrt{16}*\sqrt{2765}=4\sqrt{2765}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{2765}}{2*28}=\frac{0-4\sqrt{2765}}{56}
=-\frac{4\sqrt{2765}}{56}
=-\frac{\sqrt{2765}}{14}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{2765}}{2*28}=\frac{0+4\sqrt{2765}}{56}
=\frac{4\sqrt{2765}}{56}
=\frac{\sqrt{2765}}{14}
$