3x^2+2x=18x+4

Solution for 3x^2+2x=18x+4 equation:

3x^2+2x=18x+4

We move all terms to the left:

3x^2+2x-(18x+4)=0

We get rid of parentheses

3x^2+2x-18x-4=0

We add all the numbers together, and all the variables

3x^2-16x-4=0

a = 3; b = -16; c = -4;
Δ = b2-4ac
Δ = -162-4·3·(-4)
Δ = 304
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:

$\sqrt{\Delta}=\sqrt{304}=\sqrt{16*19}=\sqrt{16}*\sqrt{19}=4\sqrt{19}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-4\sqrt{19}}{2*3}=\frac{16-4\sqrt{19}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+4\sqrt{19}}{2*3}=\frac{16+4\sqrt{19}}{6}
$