((x-2)(x+3))=14

Solution for ((x-2)(x+3))=14 equation:

((x-2)(x+3))=14

We move all terms to the left:

((x-2)(x+3))-(14)=0

We multiply parentheses ..

((+x^2+3x-2x-6))-14=0


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

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

We get rid of parentheses

x^2+3x-2x-6

We add all the numbers together, and all the variables

x^2+x-6

Back to the equation:

+(x^2+x-6)


We get rid of parentheses

x^2+x-6-14=0

We add all the numbers together, and all the variables

x^2+x-20=0

a = 1; b = 1; c = -20;
Δ = b2-4ac
Δ = 12-4·1·(-20)
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-9}{2*1}=\frac{-10}{2}
=-5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+9}{2*1}=\frac{8}{2}
=4
$