x^2+80x+400=0

Solution for x^2+80x+400=0 equation:

x^2+80x+400=0

a = 1; b = 80; c = +400;
Δ = b2-4ac
Δ = 802-4·1·400
Δ = 4800
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:

$\sqrt{\Delta}=\sqrt{4800}=\sqrt{1600*3}=\sqrt{1600}*\sqrt{3}=40\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(80)-40\sqrt{3}}{2*1}=\frac{-80-40\sqrt{3}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(80)+40\sqrt{3}}{2*1}=\frac{-80+40\sqrt{3}}{2}
$