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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

x^2+x(x+3)-1764=0

We multiply parentheses

x^2+x^2+3x-1764=0

We add all the numbers together, and all the variables

2x^2+3x-1764=0

a = 2; b = 3; c = -1764;
Δ = b2-4ac
Δ = 32-4·2·(-1764)
Δ = 14121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{14121}=\sqrt{9*1569}=\sqrt{9}*\sqrt{1569}=3\sqrt{1569}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3\sqrt{1569}}{2*2}=\frac{-3-3\sqrt{1569}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3\sqrt{1569}}{2*2}=\frac{-3+3\sqrt{1569}}{4}
$