x^2+6x-1296=0

Solution for x^2+6x-1296=0 equation:

x^2+6x-1296=0

a = 1; b = 6; c = -1296;
Δ = b2-4ac
Δ = 62-4·1·(-1296)
Δ = 5220
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:

$\sqrt{\Delta}=\sqrt{5220}=\sqrt{36*145}=\sqrt{36}*\sqrt{145}=6\sqrt{145}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6\sqrt{145}}{2*1}=\frac{-6-6\sqrt{145}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6\sqrt{145}}{2*1}=\frac{-6+6\sqrt{145}}{2}
$