10=3x(2+x0)

Solution for 10=3x(2+x0) equation:

10=3x(2+x0)

We move all terms to the left:

10-(3x(2+x0))=0

We add all the numbers together, and all the variables

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


We calculate terms in parentheses: -(3x(x+2)), so:

3x(x+2)

We multiply parentheses

3x^2+6x

Back to the equation:

-(3x^2+6x)


We get rid of parentheses

-3x^2-6x+10=0

a = -3; b = -6; c = +10;
Δ = b2-4ac
Δ = -62-4·(-3)·10
Δ = 156
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:

$\sqrt{\Delta}=\sqrt{156}=\sqrt{4*39}=\sqrt{4}*\sqrt{39}=2\sqrt{39}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{39}}{2*-3}=\frac{6-2\sqrt{39}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{39}}{2*-3}=\frac{6+2\sqrt{39}}{-6}
$