x+3x^2=4x+12

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

x+3x^2=4x+12

We move all terms to the left:

x+3x^2-(4x+12)=0

We get rid of parentheses

3x^2+x-4x-12=0

We add all the numbers together, and all the variables

3x^2-3x-12=0

a = 3; b = -3; c = -12;
Δ = b2-4ac
Δ = -32-4·3·(-12)
Δ = 153
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:

$\sqrt{\Delta}=\sqrt{153}=\sqrt{9*17}=\sqrt{9}*\sqrt{17}=3\sqrt{17}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-3\sqrt{17}}{2*3}=\frac{3-3\sqrt{17}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3\sqrt{17}}{2*3}=\frac{3+3\sqrt{17}}{6}
$