0=(2x-3)(x+6)

Solution for 0=(2x-3)(x+6) equation:

0=(2x-3)(x+6)

We move all terms to the left:

0-((2x-3)(x+6))=0

We add all the numbers together, and all the variables

-((2x-3)(x+6))=0

We multiply parentheses ..

-((+2x^2+12x-3x-18))=0


We calculate terms in parentheses: -((+2x^2+12x-3x-18)), so:

(+2x^2+12x-3x-18)

We get rid of parentheses

2x^2+12x-3x-18

We add all the numbers together, and all the variables

2x^2+9x-18

Back to the equation:

-(2x^2+9x-18)


We get rid of parentheses

-2x^2-9x+18=0

a = -2; b = -9; c = +18;
Δ = b2-4ac
Δ = -92-4·(-2)·18
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-15}{2*-2}=\frac{-6}{-4}
=1+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+15}{2*-2}=\frac{24}{-4}
=-6
$