2x+4=0,3x^2+6x+12=12

Solution for 2x+4=0,3x^2+6x+12=12 equation:

2x+4=0.3x^2+6x+12=12

We move all terms to the left:

2x+4-(0.3x^2+6x+12)=0

We get rid of parentheses

-0.3x^2+2x-6x-12+4=0

We add all the numbers together, and all the variables

-0.3x^2-4x-8=0

a = -0.3; b = -4; c = -8;
Δ = b2-4ac
Δ = -42-4·(-0.3)·(-8)
Δ = 6.4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-\sqrt{6.4}}{2*-0.3}=\frac{4-\sqrt{6.4}}{-0.6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+\sqrt{6.4}}{2*-0.3}=\frac{4+\sqrt{6.4}}{-0.6}
$