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

Solution for (7x^2-x-4)+(x^2-2x-3)+(2x^2+3x+5)=0 equation:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

10x^2-2=0

a = 10; b = 0; c = -2;
Δ = b2-4ac
Δ = 02-4·10·(-2)
Δ = 80
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:

$\sqrt{\Delta}=\sqrt{80}=\sqrt{16*5}=\sqrt{16}*\sqrt{5}=4\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{5}}{2*10}=\frac{0-4\sqrt{5}}{20}
=-\frac{4\sqrt{5}}{20}
=-\frac{\sqrt{5}}{5}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{5}}{2*10}=\frac{0+4\sqrt{5}}{20}
=\frac{4\sqrt{5}}{20}
=\frac{\sqrt{5}}{5}
$