(2x+5)(x-4)=20

Solution for (2x+5)(x-4)=20 equation:

(2x+5)(x-4)=20

We move all terms to the left:

(2x+5)(x-4)-(20)=0

We multiply parentheses ..

(+2x^2-8x+5x-20)-20=0

We get rid of parentheses

2x^2-8x+5x-20-20=0

We add all the numbers together, and all the variables

2x^2-3x-40=0

a = 2; b = -3; c = -40;
Δ = b2-4ac
Δ = -32-4·2·(-40)
Δ = 329
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-\sqrt{329}}{2*2}=\frac{3-\sqrt{329}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{329}}{2*2}=\frac{3+\sqrt{329}}{4}
$