8a(1+5a)+4(3a-11)=20

Solution for 8a(1+5a)+4(3a-11)=20 equation:

8a(1+5a)+4(3a-11)=20

We move all terms to the left:

8a(1+5a)+4(3a-11)-(20)=0

We add all the numbers together, and all the variables

8a(5a+1)+4(3a-11)-20=0

We multiply parentheses

40a^2+8a+12a-44-20=0

We add all the numbers together, and all the variables

40a^2+20a-64=0

a = 40; b = 20; c = -64;
Δ = b2-4ac
Δ = 202-4·40·(-64)
Δ = 10640
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{10640}=\sqrt{16*665}=\sqrt{16}*\sqrt{665}=4\sqrt{665}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{665}}{2*40}=\frac{-20-4\sqrt{665}}{80}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{665}}{2*40}=\frac{-20+4\sqrt{665}}{80}
$