3(6x^2+2x-3)+2(2x^2-3x+2)=0

Solution for 3(6x^2+2x-3)+2(2x^2-3x+2)=0 equation:

3(6x^2+2x-3)+2(2x^2-3x+2)=0

We multiply parentheses

18x^2+4x^2+6x-6x-9+4=0

We add all the numbers together, and all the variables

22x^2-5=0

a = 22; b = 0; c = -5;
Δ = b2-4ac
Δ = 02-4·22·(-5)
Δ = 440
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:

$\sqrt{\Delta}=\sqrt{440}=\sqrt{4*110}=\sqrt{4}*\sqrt{110}=2\sqrt{110}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{110}}{2*22}=\frac{0-2\sqrt{110}}{44}
=-\frac{2\sqrt{110}}{44}
=-\frac{\sqrt{110}}{22}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{110}}{2*22}=\frac{0+2\sqrt{110}}{44}
=\frac{2\sqrt{110}}{44}
=\frac{\sqrt{110}}{22}
$