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

Solution for 0=(2x+5)(x-4) equation:

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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


We calculate terms in parentheses: -((+2x^2-8x+5x-20)), so:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

2x^2-3x-20

Back to the equation:

-(2x^2-3x-20)


We get rid of parentheses

-2x^2+3x+20=0

a = -2; b = 3; c = +20;
Δ = b2-4ac
Δ = 32-4·(-2)·20
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-13}{2*-2}=\frac{-16}{-4}
=+4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+13}{2*-2}=\frac{10}{-4}
=-2+1/2
$