12x^2+3x^2+20x+5=0

Solution for 12x^2+3x^2+20x+5=0 equation:

12x^2+3x^2+20x+5=0

We add all the numbers together, and all the variables

15x^2+20x+5=0

a = 15; b = 20; c = +5;
Δ = b2-4ac
Δ = 202-4·15·5
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-10}{2*15}=\frac{-30}{30}
=-1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+10}{2*15}=\frac{-10}{30}
=-1/3
$