x^2-3x-396=0

Solution for x^2-3x-396=0 equation:

x^2-3x-396=0

a = 1; b = -3; c = -396;
Δ = b2-4ac
Δ = -32-4·1·(-396)
Δ = 1593
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:

$\sqrt{\Delta}=\sqrt{1593}=\sqrt{9*177}=\sqrt{9}*\sqrt{177}=3\sqrt{177}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-3\sqrt{177}}{2*1}=\frac{3-3\sqrt{177}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3\sqrt{177}}{2*1}=\frac{3+3\sqrt{177}}{2}
$