n^2-42n+400=0

Solution for n^2-42n+400=0 equation:

n^2-42n+400=0

a = 1; b = -42; c = +400;
Δ = b2-4ac
Δ = -422-4·1·400
Δ = 164
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{164}=\sqrt{4*41}=\sqrt{4}*\sqrt{41}=2\sqrt{41}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-42)-2\sqrt{41}}{2*1}=\frac{42-2\sqrt{41}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-42)+2\sqrt{41}}{2*1}=\frac{42+2\sqrt{41}}{2}
$