(7x+1)(4x-3)=75

Solution for (7x+1)(4x-3)=75 equation:

(7x+1)(4x-3)=75

We move all terms to the left:

(7x+1)(4x-3)-(75)=0

We multiply parentheses ..

(+28x^2-21x+4x-3)-75=0

We get rid of parentheses

28x^2-21x+4x-3-75=0

We add all the numbers together, and all the variables

28x^2-17x-78=0

a = 28; b = -17; c = -78;
Δ = b2-4ac
Δ = -172-4·28·(-78)
Δ = 9025
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9025}=95$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-17)-95}{2*28}=\frac{-78}{56}
=-1+11/28
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-17)+95}{2*28}=\frac{112}{56}
=2
$