X^2+41x+390=0

Solution for X^2+41x+390=0 equation:

X^2+41X+390=0

a = 1; b = 41; c = +390;
Δ = b2-4ac
Δ = 412-4·1·390
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(41)-11}{2*1}=\frac{-52}{2}
=-26
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(41)+11}{2*1}=\frac{-30}{2}
=-15
$