x^2-4x=405

Solution for x^2-4x=405 equation:

x^2-4x=405

We move all terms to the left:

x^2-4x-(405)=0

a = 1; b = -4; c = -405;
Δ = b2-4ac
Δ = -42-4·1·(-405)
Δ = 1636
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:

$\sqrt{\Delta}=\sqrt{1636}=\sqrt{4*409}=\sqrt{4}*\sqrt{409}=2\sqrt{409}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2\sqrt{409}}{2*1}=\frac{4-2\sqrt{409}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2\sqrt{409}}{2*1}=\frac{4+2\sqrt{409}}{2}
$