2(3x+5)=6(x-2);x=

Solution for 2(3x+5)=6(x-2);x= equation:

2(3x+5)=6(x-2)x=

We move all terms to the left:

2(3x+5)-(6(x-2)x)=0

We multiply parentheses

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


We calculate terms in parentheses: -(6(x-2)x), so:

6(x-2)x

We multiply parentheses

6x^2-12x

Back to the equation:

-(6x^2-12x)


We get rid of parentheses

-6x^2+6x+12x+10=0

We add all the numbers together, and all the variables

-6x^2+18x+10=0

a = -6; b = 18; c = +10;
Δ = b2-4ac
Δ = 182-4·(-6)·10
Δ = 564
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:

$\sqrt{\Delta}=\sqrt{564}=\sqrt{4*141}=\sqrt{4}*\sqrt{141}=2\sqrt{141}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-2\sqrt{141}}{2*-6}=\frac{-18-2\sqrt{141}}{-12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+2\sqrt{141}}{2*-6}=\frac{-18+2\sqrt{141}}{-12}
$