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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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


We calculate terms in parentheses: -(-(+x^2-2x+3x-6)), so:

-(+x^2-2x+3x-6)

We get rid of parentheses

-x^2+2x-3x+6

We add all the numbers together, and all the variables

-1x^2-1x+6

Back to the equation:

-(-1x^2-1x+6)


We get rid of parentheses

1x^2+1x-6=0

We add all the numbers together, and all the variables

x^2+x-6=0

a = 1; b = 1; c = -6;
Δ = b2-4ac
Δ = 12-4·1·(-6)
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-5}{2*1}=\frac{-6}{2}
=-3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+5}{2*1}=\frac{4}{2}
=2
$