2x(3+x)=10x+1680

Solution for 2x(3+x)=10x+1680 equation:

2x(3+x)=10x+1680

We move all terms to the left:

2x(3+x)-(10x+1680)=0

We add all the numbers together, and all the variables

2x(x+3)-(10x+1680)=0

We multiply parentheses

2x^2+6x-(10x+1680)=0

We get rid of parentheses

2x^2+6x-10x-1680=0

We add all the numbers together, and all the variables

2x^2-4x-1680=0

a = 2; b = -4; c = -1680;
Δ = b2-4ac
Δ = -42-4·2·(-1680)
Δ = 13456
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{13456}=116$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-116}{2*2}=\frac{-112}{4}
=-28
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+116}{2*2}=\frac{120}{4}
=30
$