12x^2+8x-407=0

Solution for 12x^2+8x-407=0 equation:

12x^2+8x-407=0

a = 12; b = 8; c = -407;
Δ = b2-4ac
Δ = 82-4·12·(-407)
Δ = 19600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{19600}=140$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-140}{2*12}=\frac{-148}{24}
=-6+1/6
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+140}{2*12}=\frac{132}{24}
=5+1/2
$