6x2-3x=3(7x2-4x)

Solution for 6x2-3x=3(7x2-4x) equation:

6x^2-3x=3(7x^2-4x)

We move all terms to the left:

6x^2-3x-(3(7x^2-4x))=0


We calculate terms in parentheses: -(3(7x^2-4x)), so:

3(7x^2-4x)

We multiply parentheses

21x^2-12x

Back to the equation:

-(21x^2-12x)


We get rid of parentheses

6x^2-21x^2-3x+12x=0

We add all the numbers together, and all the variables

-15x^2+9x=0

a = -15; b = 9; c = 0;
Δ = b2-4ac
Δ = 92-4·(-15)·0
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-9}{2*-15}=\frac{-18}{-30}
=3/5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+9}{2*-15}=\frac{0}{-30}
=0
$