0=(2x-3)(x+4)

Solution for 0=(2x-3)(x+4) equation:

0=(2x-3)(x+4)

We move all terms to the left:

0-((2x-3)(x+4))=0

We add all the numbers together, and all the variables

-((2x-3)(x+4))=0

We multiply parentheses ..

-((+2x^2+8x-3x-12))=0


We calculate terms in parentheses: -((+2x^2+8x-3x-12)), so:

(+2x^2+8x-3x-12)

We get rid of parentheses

2x^2+8x-3x-12

We add all the numbers together, and all the variables

2x^2+5x-12

Back to the equation:

-(2x^2+5x-12)


We get rid of parentheses

-2x^2-5x+12=0

a = -2; b = -5; c = +12;
Δ = b2-4ac
Δ = -52-4·(-2)·12
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-11}{2*-2}=\frac{-6}{-4}
=1+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+11}{2*-2}=\frac{16}{-4}
=-4
$