(2x+5)(x+4)=1

Solution for (2x+5)(x+4)=1 equation:

(2x+5)(x+4)=1

We move all terms to the left:

(2x+5)(x+4)-(1)=0

We multiply parentheses ..

(+2x^2+8x+5x+20)-1=0

We get rid of parentheses

2x^2+8x+5x+20-1=0

We add all the numbers together, and all the variables

2x^2+13x+19=0

a = 2; b = 13; c = +19;
Δ = b2-4ac
Δ = 132-4·2·19
Δ = 17
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{-(13)-\sqrt{17}}{2*2}=\frac{-13-\sqrt{17}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+\sqrt{17}}{2*2}=\frac{-13+\sqrt{17}}{4}
$