(7x^2-4)+(x^2+3)+(3x2+5)+(5x^2-2)=67

Solution for (7x^2-4)+(x^2+3)+(3x2+5)+(5x^2-2)=67 equation:

(7x^2-4)+(x^2+3)+(3x^2+5)+(5x^2-2)=67

We move all terms to the left:

(7x^2-4)+(x^2+3)+(3x^2+5)+(5x^2-2)-(67)=0

We get rid of parentheses

7x^2+x^2+3x^2+5x^2-4+3+5-2-67=0

We add all the numbers together, and all the variables

16x^2-65=0

a = 16; b = 0; c = -65;
Δ = b2-4ac
Δ = 02-4·16·(-65)
Δ = 4160
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:

$\sqrt{\Delta}=\sqrt{4160}=\sqrt{64*65}=\sqrt{64}*\sqrt{65}=8\sqrt{65}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{65}}{2*16}=\frac{0-8\sqrt{65}}{32}
=-\frac{8\sqrt{65}}{32}
=-\frac{\sqrt{65}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{65}}{2*16}=\frac{0+8\sqrt{65}}{32}
=\frac{8\sqrt{65}}{32}
=\frac{\sqrt{65}}{4}
$