3(2x2-9+6x2)=3(2)+29-7(2)

Solution for 3(2x2-9+6x2)=3(2)+29-7(2) equation:

3(2x^2-9+6x^2)=3(2)+29-7(2)

We move all terms to the left:

3(2x^2-9+6x^2)-(3(2)+29-7(2))=0

We add all the numbers together, and all the variables

3(2x^2-9+6x^2)-(-11)=0

We add all the numbers together, and all the variables

3(2x^2-9+6x^2)+11=0

We multiply parentheses

6x^2+18x^2-27+11=0

We add all the numbers together, and all the variables

24x^2-16=0

a = 24; b = 0; c = -16;
Δ = b2-4ac
Δ = 02-4·24·(-16)
Δ = 1536
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:

$\sqrt{\Delta}=\sqrt{1536}=\sqrt{256*6}=\sqrt{256}*\sqrt{6}=16\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{6}}{2*24}=\frac{0-16\sqrt{6}}{48}
=-\frac{16\sqrt{6}}{48}
=-\frac{\sqrt{6}}{3}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{6}}{2*24}=\frac{0+16\sqrt{6}}{48}
=\frac{16\sqrt{6}}{48}
=\frac{\sqrt{6}}{3}
$