(x+3)+(x^2-9)=0

Solution for (x+3)+(x^2-9)=0 equation:

(x+3)+(x^2-9)=0

We get rid of parentheses

x^2+x+3-9=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
$