2(3x2-22)+2(9x2+20)=400

Solution for 2(3x2-22)+2(9x2+20)=400 equation:

2(3x^2-22)+2(9x^2+20)=400

We move all terms to the left:

2(3x^2-22)+2(9x^2+20)-(400)=0

We multiply parentheses

6x^2+18x^2-44+40-400=0

We add all the numbers together, and all the variables

24x^2-404=0

a = 24; b = 0; c = -404;
Δ = b2-4ac
Δ = 02-4·24·(-404)
Δ = 38784
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:

$\sqrt{\Delta}=\sqrt{38784}=\sqrt{64*606}=\sqrt{64}*\sqrt{606}=8\sqrt{606}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{606}}{2*24}=\frac{0-8\sqrt{606}}{48}
=-\frac{8\sqrt{606}}{48}
=-\frac{\sqrt{606}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{606}}{2*24}=\frac{0+8\sqrt{606}}{48}
=\frac{8\sqrt{606}}{48}
=\frac{\sqrt{606}}{6}
$