2x(3x+2x)=312

Solution for 2x(3x+2x)=312 equation:

2x(3x+2x)=312

We move all terms to the left:

2x(3x+2x)-(312)=0

We add all the numbers together, and all the variables

2x(+5x)-312=0

We multiply parentheses

10x^2-312=0

a = 10; b = 0; c = -312;
Δ = b2-4ac
Δ = 02-4·10·(-312)
Δ = 12480
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:

$\sqrt{\Delta}=\sqrt{12480}=\sqrt{64*195}=\sqrt{64}*\sqrt{195}=8\sqrt{195}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{195}}{2*10}=\frac{0-8\sqrt{195}}{20}
=-\frac{8\sqrt{195}}{20}
=-\frac{2\sqrt{195}}{5}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{195}}{2*10}=\frac{0+8\sqrt{195}}{20}
=\frac{8\sqrt{195}}{20}
=\frac{2\sqrt{195}}{5}
$