28x^2-4x=6(5x^2+2x)

Solution for 28x^2-4x=6(5x^2+2x) equation:

28x^2-4x=6(5x^2+2x)

We move all terms to the left:

28x^2-4x-(6(5x^2+2x))=0


We calculate terms in parentheses: -(6(5x^2+2x)), so:

6(5x^2+2x)

We multiply parentheses

30x^2+12x

Back to the equation:

-(30x^2+12x)


We get rid of parentheses

28x^2-30x^2-4x-12x=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{0}{-4}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+16}{2*-2}=\frac{32}{-4}
=-8
$