2x(3x+35)=(x-5)

Solution for 2x(3x+35)=(x-5) equation:

2x(3x+35)=(x-5)

We move all terms to the left:

2x(3x+35)-((x-5))=0

We multiply parentheses

6x^2+70x-((x-5))=0


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

(x-5)

We get rid of parentheses

x-5

Back to the equation:

-(x-5)


We get rid of parentheses

6x^2+70x-x+5=0

We add all the numbers together, and all the variables

6x^2+69x+5=0

a = 6; b = 69; c = +5;
Δ = b2-4ac
Δ = 692-4·6·5
Δ = 4641
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(69)-\sqrt{4641}}{2*6}=\frac{-69-\sqrt{4641}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(69)+\sqrt{4641}}{2*6}=\frac{-69+\sqrt{4641}}{12}
$