13(2x)+20=6x(5x+2)

Solution for 13(2x)+20=6x(5x+2) equation:

13(2x)+20=6x(5x+2)

We move all terms to the left:

13(2x)+20-(6x(5x+2))=0


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

6x(5x+2)

We multiply parentheses

30x^2+12x

Back to the equation:

-(30x^2+12x)


We get rid of parentheses

-30x^2+132x-12x+20=0

We add all the numbers together, and all the variables

-30x^2+120x+20=0

a = -30; b = 120; c = +20;
Δ = b2-4ac
Δ = 1202-4·(-30)·20
Δ = 16800
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{16800}=\sqrt{400*42}=\sqrt{400}*\sqrt{42}=20\sqrt{42}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(120)-20\sqrt{42}}{2*-30}=\frac{-120-20\sqrt{42}}{-60}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(120)+20\sqrt{42}}{2*-30}=\frac{-120+20\sqrt{42}}{-60}
$