0=(3x-7)(x+9)

Solution for 0=(3x-7)(x+9) equation:

0=(3x-7)(x+9)

We move all terms to the left:

0-((3x-7)(x+9))=0

We add all the numbers together, and all the variables

-((3x-7)(x+9))=0

We multiply parentheses ..

-((+3x^2+27x-7x-63))=0


We calculate terms in parentheses: -((+3x^2+27x-7x-63)), so:

(+3x^2+27x-7x-63)

We get rid of parentheses

3x^2+27x-7x-63

We add all the numbers together, and all the variables

3x^2+20x-63

Back to the equation:

-(3x^2+20x-63)


We get rid of parentheses

-3x^2-20x+63=0

a = -3; b = -20; c = +63;
Δ = b2-4ac
Δ = -202-4·(-3)·63
Δ = 1156
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1156}=34$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-34}{2*-3}=\frac{-14}{-6}
=2+1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+34}{2*-3}=\frac{54}{-6}
=-9
$