x^2-5x^2+20+6(x^2+1+8x)=0

Solution for x^2-5x^2+20+6(x^2+1+8x)=0 equation:

x^2-5x^2+20+6(x^2+1+8x)=0

We add all the numbers together, and all the variables

-4x^2+6(x^2+1+8x)+20=0

We multiply parentheses

-4x^2+6x^2+48x+6+20=0

We add all the numbers together, and all the variables

2x^2+48x+26=0

a = 2; b = 48; c = +26;
Δ = b2-4ac
Δ = 482-4·2·26
Δ = 2096
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:

$\sqrt{\Delta}=\sqrt{2096}=\sqrt{16*131}=\sqrt{16}*\sqrt{131}=4\sqrt{131}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-4\sqrt{131}}{2*2}=\frac{-48-4\sqrt{131}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+4\sqrt{131}}{2*2}=\frac{-48+4\sqrt{131}}{4}
$