3(2x^2+4)=27

Solution for 3(2x^2+4)=27 equation:

3(2x^2+4)=27

We move all terms to the left:

3(2x^2+4)-(27)=0

We multiply parentheses

6x^2+12-27=0

We add all the numbers together, and all the variables

6x^2-15=0

a = 6; b = 0; c = -15;
Δ = b2-4ac
Δ = 02-4·6·(-15)
Δ = 360
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:

$\sqrt{\Delta}=\sqrt{360}=\sqrt{36*10}=\sqrt{36}*\sqrt{10}=6\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{10}}{2*6}=\frac{0-6\sqrt{10}}{12}
=-\frac{6\sqrt{10}}{12}
=-\frac{\sqrt{10}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{10}}{2*6}=\frac{0+6\sqrt{10}}{12}
=\frac{6\sqrt{10}}{12}
=\frac{\sqrt{10}}{2}
$