x(x+3)+7+4x+(3x-6)=40

Solution for x(x+3)+7+4x+(3x-6)=40 equation:

x(x+3)+7+4x+(3x-6)=40

We move all terms to the left:

x(x+3)+7+4x+(3x-6)-(40)=0

We add all the numbers together, and all the variables

4x+x(x+3)+(3x-6)-33=0

We multiply parentheses

x^2+4x+3x+(3x-6)-33=0

We get rid of parentheses

x^2+4x+3x+3x-6-33=0

We add all the numbers together, and all the variables

x^2+10x-39=0

a = 1; b = 10; c = -39;
Δ = b2-4ac
Δ = 102-4·1·(-39)
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-16}{2*1}=\frac{-26}{2}
=-13
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+16}{2*1}=\frac{6}{2}
=3
$