3x2+2(x2-20)=16-2x2

Solution for 3x2+2(x2-20)=16-2x2 equation:

3x^2+2(x2-20)=16-2x^2

We move all terms to the left:

3x^2+2(x2-20)-(16-2x^2)=0

We add all the numbers together, and all the variables

3x^2-(16-2x^2)+2(+x^2-20)=0

We multiply parentheses

3x^2-(16-2x^2)+2x^2-40=0

We get rid of parentheses

3x^2+2x^2+2x^2-16-40=0

We add all the numbers together, and all the variables

7x^2-56=0

a = 7; b = 0; c = -56;
Δ = b2-4ac
Δ = 02-4·7·(-56)
Δ = 1568
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:

$\sqrt{\Delta}=\sqrt{1568}=\sqrt{784*2}=\sqrt{784}*\sqrt{2}=28\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-28\sqrt{2}}{2*7}=\frac{0-28\sqrt{2}}{14}
=-\frac{28\sqrt{2}}{14}
=-2\sqrt{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+28\sqrt{2}}{2*7}=\frac{0+28\sqrt{2}}{14}
=\frac{28\sqrt{2}}{14}
=2\sqrt{2}
$