x2-1490x+256=0

Solution for x2-1490x+256=0 equation:

x2-1490x+256=0

We add all the numbers together, and all the variables

x^2-1490x+256=0

a = 1; b = -1490; c = +256;
Δ = b2-4ac
Δ = -14902-4·1·256
Δ = 2219076
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:

$\sqrt{\Delta}=\sqrt{2219076}=\sqrt{2916*761}=\sqrt{2916}*\sqrt{761}=54\sqrt{761}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1490)-54\sqrt{761}}{2*1}=\frac{1490-54\sqrt{761}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1490)+54\sqrt{761}}{2*1}=\frac{1490+54\sqrt{761}}{2}
$