X=3x^2+13x+4

Solution for X=3x^2+13x+4 equation:

X=3X^2+13X+4

We move all terms to the left:

X-(3X^2+13X+4)=0

We get rid of parentheses

-3X^2+X-13X-4=0

We add all the numbers together, and all the variables

-3X^2-12X-4=0

a = -3; b = -12; c = -4;
Δ = b2-4ac
Δ = -122-4·(-3)·(-4)
Δ = 96
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:

$\sqrt{\Delta}=\sqrt{96}=\sqrt{16*6}=\sqrt{16}*\sqrt{6}=4\sqrt{6}$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4\sqrt{6}}{2*-3}=\frac{12-4\sqrt{6}}{-6}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{6}}{2*-3}=\frac{12+4\sqrt{6}}{-6}
$