x^2+16-320=0

Solution for x^2+16-320=0 equation:

x^2+16-320=0

We add all the numbers together, and all the variables

x^2-304=0

a = 1; b = 0; c = -304;
Δ = b2-4ac
Δ = 02-4·1·(-304)
Δ = 1216
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:

$\sqrt{\Delta}=\sqrt{1216}=\sqrt{64*19}=\sqrt{64}*\sqrt{19}=8\sqrt{19}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{19}}{2*1}=\frac{0-8\sqrt{19}}{2}
=-\frac{8\sqrt{19}}{2}
=-4\sqrt{19}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{19}}{2*1}=\frac{0+8\sqrt{19}}{2}
=\frac{8\sqrt{19}}{2}
=4\sqrt{19}
$