12x^2+20x+4=0

Solution for 12x^2+20x+4=0 equation:

12x^2+20x+4=0

a = 12; b = 20; c = +4;
Δ = b2-4ac
Δ = 202-4·12·4
Δ = 208
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:

$\sqrt{\Delta}=\sqrt{208}=\sqrt{16*13}=\sqrt{16}*\sqrt{13}=4\sqrt{13}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{13}}{2*12}=\frac{-20-4\sqrt{13}}{24}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{13}}{2*12}=\frac{-20+4\sqrt{13}}{24}
$