(x+2)(x+3)+9x+42=0

Solution for (x+2)(x+3)+9x+42=0 equation:

(x+2)(x+3)+9x+42=0

We add all the numbers together, and all the variables

9x+(x+2)(x+3)+42=0

We multiply parentheses ..

(+x^2+3x+2x+6)+9x+42=0

We get rid of parentheses

x^2+3x+2x+9x+6+42=0

We add all the numbers together, and all the variables

x^2+14x+48=0

a = 1; b = 14; c = +48;
Δ = b2-4ac
Δ = 142-4·1·48
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-2}{2*1}=\frac{-16}{2}
=-8
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+2}{2*1}=\frac{-12}{2}
=-6
$