(7x^2-4)+(x^2+3)+(3x2+5)+(5x^2-2)=66

Solution for (7x^2-4)+(x^2+3)+(3x2+5)+(5x^2-2)=66 equation:

(7x^2-4)+(x^2+3)+(3x^2+5)+(5x^2-2)=66

We move all terms to the left:

(7x^2-4)+(x^2+3)+(3x^2+5)+(5x^2-2)-(66)=0

We get rid of parentheses

7x^2+x^2+3x^2+5x^2-4+3+5-2-66=0

We add all the numbers together, and all the variables

16x^2-64=0

a = 16; b = 0; c = -64;
Δ = b2-4ac
Δ = 02-4·16·(-64)
Δ = 4096
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4096}=64$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-64}{2*16}=\frac{-64}{32}
=-2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+64}{2*16}=\frac{64}{32}
=2
$