12+14x+4x^2=20

Solution for 12+14x+4x^2=20 equation:

12+14x+4x^2=20

We move all terms to the left:

12+14x+4x^2-(20)=0

We add all the numbers together, and all the variables

4x^2+14x-8=0

a = 4; b = 14; c = -8;
Δ = b2-4ac
Δ = 142-4·4·(-8)
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-18}{2*4}=\frac{-32}{8}
=-4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+18}{2*4}=\frac{4}{8}
=1/2
$