2x2+(x-8)(x+8)-3(1-x2)=0

Solution for 2x2+(x-8)(x+8)-3(1-x2)=0 equation:

2x^2+(x-8)(x+8)-3(1-x2)=0

We add all the numbers together, and all the variables

2x^2-3(-1x^2+1)+(x-8)(x+8)=0

We use the square of the difference formula

2x^2-3(-1x^2+1)+x^2-64=0

We multiply parentheses

2x^2+3x^2+x^2-3-64=0

We add all the numbers together, and all the variables

6x^2-67=0

a = 6; b = 0; c = -67;
Δ = b2-4ac
Δ = 02-4·6·(-67)
Δ = 1608
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:

$\sqrt{\Delta}=\sqrt{1608}=\sqrt{4*402}=\sqrt{4}*\sqrt{402}=2\sqrt{402}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{402}}{2*6}=\frac{0-2\sqrt{402}}{12}
=-\frac{2\sqrt{402}}{12}
=-\frac{\sqrt{402}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{402}}{2*6}=\frac{0+2\sqrt{402}}{12}
=\frac{2\sqrt{402}}{12}
=\frac{\sqrt{402}}{6}
$