12x^2+12x-419=0

Solution for 12x^2+12x-419=0 equation:

12x^2+12x-419=0

a = 12; b = 12; c = -419;
Δ = b2-4ac
Δ = 122-4·12·(-419)
Δ = 20256
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:

$\sqrt{\Delta}=\sqrt{20256}=\sqrt{16*1266}=\sqrt{16}*\sqrt{1266}=4\sqrt{1266}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4\sqrt{1266}}{2*12}=\frac{-12-4\sqrt{1266}}{24}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4\sqrt{1266}}{2*12}=\frac{-12+4\sqrt{1266}}{24}
$