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

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

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

We move all terms to the left:

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

We multiply parentheses

30x^2-12x-((2x-3)(5x-2))=0

We multiply parentheses ..

30x^2-((+10x^2-4x-15x+6))-12x=0


We calculate terms in parentheses: -((+10x^2-4x-15x+6)), so:

(+10x^2-4x-15x+6)

We get rid of parentheses

10x^2-4x-15x+6

We add all the numbers together, and all the variables

10x^2-19x+6

Back to the equation:

-(10x^2-19x+6)


We add all the numbers together, and all the variables

30x^2-12x-(10x^2-19x+6)=0

We get rid of parentheses

30x^2-10x^2-12x+19x-6=0

We add all the numbers together, and all the variables

20x^2+7x-6=0

a = 20; b = 7; c = -6;
Δ = b2-4ac
Δ = 72-4·20·(-6)
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-23}{2*20}=\frac{-30}{40}
=-3/4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+23}{2*20}=\frac{16}{40}
=2/5
$