x^2+82x+40=0

Solution for x^2+82x+40=0 equation:

x^2+82x+40=0

a = 1; b = 82; c = +40;
Δ = b2-4ac
Δ = 822-4·1·40
Δ = 6564
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:

$\sqrt{\Delta}=\sqrt{6564}=\sqrt{4*1641}=\sqrt{4}*\sqrt{1641}=2\sqrt{1641}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(82)-2\sqrt{1641}}{2*1}=\frac{-82-2\sqrt{1641}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(82)+2\sqrt{1641}}{2*1}=\frac{-82+2\sqrt{1641}}{2}
$