(5x)+(5x)+(11x+3)+(11x+3)x=4

Solution for (5x)+(5x)+(11x+3)+(11x+3)x=4 equation:

(5x)+(5x)+(11x+3)+(11x+3)x=4

We move all terms to the left:

(5x)+(5x)+(11x+3)+(11x+3)x-(4)=0

We add all the numbers together, and all the variables

10x+(11x+3)+(11x+3)x-4=0

We multiply parentheses

11x^2+10x+(11x+3)+3x-4=0

We get rid of parentheses

11x^2+10x+11x+3x+3-4=0

We add all the numbers together, and all the variables

11x^2+24x-1=0

a = 11; b = 24; c = -1;
Δ = b2-4ac
Δ = 242-4·11·(-1)
Δ = 620
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:

$\sqrt{\Delta}=\sqrt{620}=\sqrt{4*155}=\sqrt{4}*\sqrt{155}=2\sqrt{155}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-2\sqrt{155}}{2*11}=\frac{-24-2\sqrt{155}}{22}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+2\sqrt{155}}{2*11}=\frac{-24+2\sqrt{155}}{22}
$