x^2+42x+80=0

Solution for x^2+42x+80=0 equation:

x^2+42x+80=0

a = 1; b = 42; c = +80;
Δ = b2-4ac
Δ = 422-4·1·80
Δ = 1444
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1444}=38$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-38}{2*1}=\frac{-80}{2}
=-40
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+38}{2*1}=\frac{-4}{2}
=-2
$