(x^2)+(x+1^2)-256=0

Solution for (x^2)+(x+1^2)-256=0 equation:

(x^2)+(x+1^2)-256=0

We get rid of parentheses

x^2+x-256+1^2=0

We add all the numbers together, and all the variables

x^2+x-255=0

a = 1; b = 1; c = -255;
Δ = b2-4ac
Δ = 12-4·1·(-255)
Δ = 1021
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{1021}}{2*1}=\frac{-1-\sqrt{1021}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{1021}}{2*1}=\frac{-1+\sqrt{1021}}{2}
$