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

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

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

We move all terms to the left:

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

We multiply parentheses ..

(+3x^2-15x-2x+10)-(2(x+3)-8)=0


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

2(x+3)-8

We multiply parentheses

2x+6-8

We add all the numbers together, and all the variables

2x-2

Back to the equation:

-(2x-2)


We get rid of parentheses

3x^2-15x-2x-2x+10+2=0

We add all the numbers together, and all the variables

3x^2-19x+12=0

a = 3; b = -19; c = +12;
Δ = b2-4ac
Δ = -192-4·3·12
Δ = 217
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-\sqrt{217}}{2*3}=\frac{19-\sqrt{217}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+\sqrt{217}}{2*3}=\frac{19+\sqrt{217}}{6}
$