x^2+x-3906=0

Solution for x^2+x-3906=0 equation:

x^2+x-3906=0

a = 1; b = 1; c = -3906;
Δ = b2-4ac
Δ = 12-4·1·(-3906)
Δ = 15625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{15625}=125$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-125}{2*1}=\frac{-126}{2}
=-63
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+125}{2*1}=\frac{124}{2}
=62
$