13x^2-3x=2(x-7x^2)

Solution for 13x^2-3x=2(x-7x^2) equation:

13x^2-3x=2(x-7x^2)

We move all terms to the left:

13x^2-3x-(2(x-7x^2))=0


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

2(x-7x^2)

We multiply parentheses

-14x^2+2x

Back to the equation:

-(-14x^2+2x)


We get rid of parentheses

13x^2+14x^2-2x-3x=0

We add all the numbers together, and all the variables

27x^2-5x=0

a = 27; b = -5; c = 0;
Δ = b2-4ac
Δ = -52-4·27·0
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-5}{2*27}=\frac{0}{54}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+5}{2*27}=\frac{10}{54}
=5/27
$