3x-2(x-5)=2x(x+3)-8

Solution for 3x-2(x-5)=2x(x+3)-8 equation:

3x-2(x-5)=2x(x+3)-8

We move all terms to the left:

3x-2(x-5)-(2x(x+3)-8)=0

We multiply parentheses

3x-2x-(2x(x+3)-8)+10=0


We calculate terms in parentheses: -(2x(x+3)-8), so:

2x(x+3)-8

We multiply parentheses

2x^2+6x-8

Back to the equation:

-(2x^2+6x-8)


We add all the numbers together, and all the variables

x-(2x^2+6x-8)+10=0

We get rid of parentheses

-2x^2+x-6x+8+10=0

We add all the numbers together, and all the variables

-2x^2-5x+18=0

a = -2; b = -5; c = +18;
Δ = b2-4ac
Δ = -52-4·(-2)·18
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-13}{2*-2}=\frac{-8}{-4}
=+2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+13}{2*-2}=\frac{18}{-4}
=-4+1/2
$