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

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

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

We move all terms to the left:

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

We multiply parentheses

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

We add all the numbers together, and all the variables

2x^2+3x-42=0

a = 2; b = 3; c = -42;
Δ = b2-4ac
Δ = 32-4·2·(-42)
Δ = 345
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{345}}{2*2}=\frac{-3-\sqrt{345}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{345}}{2*2}=\frac{-3+\sqrt{345}}{4}
$