x(3+x)=1120

Solution for x(3+x)=1120 equation:

x(3+x)=1120

We move all terms to the left:

x(3+x)-(1120)=0

We add all the numbers together, and all the variables

x(x+3)-1120=0

We multiply parentheses

x^2+3x-1120=0

a = 1; b = 3; c = -1120;
Δ = b2-4ac
Δ = 32-4·1·(-1120)
Δ = 4489
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4489}=67$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-67}{2*1}=\frac{-70}{2}
=-35
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+67}{2*1}=\frac{64}{2}
=32
$