x(x+1)+(x+2)+(x+3)=42

Solution for x(x+1)+(x+2)+(x+3)=42 equation:

x(x+1)+(x+2)+(x+3)=42

We move all terms to the left:

x(x+1)+(x+2)+(x+3)-(42)=0

We multiply parentheses

x^2+x+(x+2)+(x+3)-42=0

We get rid of parentheses

x^2+x+x+x+2+3-42=0

We add all the numbers together, and all the variables

x^2+3x-37=0

a = 1; b = 3; c = -37;
Δ = b2-4ac
Δ = 32-4·1·(-37)
Δ = 157
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-\sqrt{157}}{2*1}=\frac{-3-\sqrt{157}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{157}}{2*1}=\frac{-3+\sqrt{157}}{2}
$