(7x+1)(4x-1)=8

Solution for (7x+1)(4x-1)=8 equation:

(7x+1)(4x-1)=8

We move all terms to the left:

(7x+1)(4x-1)-(8)=0

We multiply parentheses ..

(+28x^2-7x+4x-1)-8=0

We get rid of parentheses

28x^2-7x+4x-1-8=0

We add all the numbers together, and all the variables

28x^2-3x-9=0

a = 28; b = -3; c = -9;
Δ = b2-4ac
Δ = -32-4·28·(-9)
Δ = 1017
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:

$\sqrt{\Delta}=\sqrt{1017}=\sqrt{9*113}=\sqrt{9}*\sqrt{113}=3\sqrt{113}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-3\sqrt{113}}{2*28}=\frac{3-3\sqrt{113}}{56}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3\sqrt{113}}{2*28}=\frac{3+3\sqrt{113}}{56}
$