(2x+1)(x-5)=40

Solution for (2x+1)(x-5)=40 equation:

(2x+1)(x-5)=40

We move all terms to the left:

(2x+1)(x-5)-(40)=0

We multiply parentheses ..

(+2x^2-10x+x-5)-40=0

We get rid of parentheses

2x^2-10x+x-5-40=0

We add all the numbers together, and all the variables

2x^2-9x-45=0

a = 2; b = -9; c = -45;
Δ = b2-4ac
Δ = -92-4·2·(-45)
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-21}{2*2}=\frac{-12}{4}
=-3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+21}{2*2}=\frac{30}{4}
=7+1/2
$