2x(3+x)=16x+1500

Solution for 2x(3+x)=16x+1500 equation:

2x(3+x)=16x+1500

We move all terms to the left:

2x(3+x)-(16x+1500)=0

We add all the numbers together, and all the variables

2x(x+3)-(16x+1500)=0

We multiply parentheses

2x^2+6x-(16x+1500)=0

We get rid of parentheses

2x^2+6x-16x-1500=0

We add all the numbers together, and all the variables

2x^2-10x-1500=0

a = 2; b = -10; c = -1500;
Δ = b2-4ac
Δ = -102-4·2·(-1500)
Δ = 12100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{12100}=110$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-110}{2*2}=\frac{-100}{4}
=-25
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+110}{2*2}=\frac{120}{4}
=30
$