(x+1)(4x-5)=100

Solution for (x+1)(4x-5)=100 equation:

(x+1)(4x-5)=100

We move all terms to the left:

(x+1)(4x-5)-(100)=0

We multiply parentheses ..

(+4x^2-5x+4x-5)-100=0

We get rid of parentheses

4x^2-5x+4x-5-100=0

We add all the numbers together, and all the variables

4x^2-1x-105=0

a = 4; b = -1; c = -105;
Δ = b2-4ac
Δ = -12-4·4·(-105)
Δ = 1681
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1681}=41$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-41}{2*4}=\frac{-40}{8}
=-5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+41}{2*4}=\frac{42}{8}
=5+1/4
$