2x(3x+2x)=650

Solution for 2x(3x+2x)=650 equation:

2x(3x+2x)=650

We move all terms to the left:

2x(3x+2x)-(650)=0

We add all the numbers together, and all the variables

2x(+5x)-650=0

We multiply parentheses

10x^2-650=0

a = 10; b = 0; c = -650;
Δ = b2-4ac
Δ = 02-4·10·(-650)
Δ = 26000
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:

$\sqrt{\Delta}=\sqrt{26000}=\sqrt{400*65}=\sqrt{400}*\sqrt{65}=20\sqrt{65}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-20\sqrt{65}}{2*10}=\frac{0-20\sqrt{65}}{20}
=-\frac{20\sqrt{65}}{20}
=-\sqrt{65}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+20\sqrt{65}}{2*10}=\frac{0+20\sqrt{65}}{20}
=\frac{20\sqrt{65}}{20}
=\sqrt{65}
$