(13+2x)(4x+1)=3+4

Solution for (13+2x)(4x+1)=3+4 equation:

(13+2x)(4x+1)=3+4

We move all terms to the left:

(13+2x)(4x+1)-(3+4)=0

We add all the numbers together, and all the variables

(2x+13)(4x+1)-7=0

We multiply parentheses ..

(+8x^2+2x+52x+13)-7=0

We get rid of parentheses

8x^2+2x+52x+13-7=0

We add all the numbers together, and all the variables

8x^2+54x+6=0

a = 8; b = 54; c = +6;
Δ = b2-4ac
Δ = 542-4·8·6
Δ = 2724
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:

$\sqrt{\Delta}=\sqrt{2724}=\sqrt{4*681}=\sqrt{4}*\sqrt{681}=2\sqrt{681}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(54)-2\sqrt{681}}{2*8}=\frac{-54-2\sqrt{681}}{16}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(54)+2\sqrt{681}}{2*8}=\frac{-54+2\sqrt{681}}{16}
$