(5x+1)(x+2)=(x-1)(3x-2)

Solution for (5x+1)(x+2)=(x-1)(3x-2) equation:

(5x+1)(x+2)=(x-1)(3x-2)

We move all terms to the left:

(5x+1)(x+2)-((x-1)(3x-2))=0

We multiply parentheses ..

(+5x^2+10x+x+2)-((x-1)(3x-2))=0


We calculate terms in parentheses: -((x-1)(3x-2)), so:

(x-1)(3x-2)

We multiply parentheses ..

(+3x^2-2x-3x+2)

We get rid of parentheses

3x^2-2x-3x+2

We add all the numbers together, and all the variables

3x^2-5x+2

Back to the equation:

-(3x^2-5x+2)


We get rid of parentheses

5x^2-3x^2+10x+x+5x+2-2=0

We add all the numbers together, and all the variables

2x^2+16x=0

a = 2; b = 16; c = 0;
Δ = b2-4ac
Δ = 162-4·2·0
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-16}{2*2}=\frac{-32}{4}
=-8
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+16}{2*2}=\frac{0}{4}
=0
$