3x(2x-7)=(2x+35)

Solution for 3x(2x-7)=(2x+35) equation:

3x(2x-7)=(2x+35)

We move all terms to the left:

3x(2x-7)-((2x+35))=0

We multiply parentheses

6x^2-21x-((2x+35))=0


We calculate terms in parentheses: -((2x+35)), so:

(2x+35)

We get rid of parentheses

2x+35

Back to the equation:

-(2x+35)


We get rid of parentheses

6x^2-21x-2x-35=0

We add all the numbers together, and all the variables

6x^2-23x-35=0

a = 6; b = -23; c = -35;
Δ = b2-4ac
Δ = -232-4·6·(-35)
Δ = 1369
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1369}=37$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-23)-37}{2*6}=\frac{-14}{12}
=-1+1/6
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+37}{2*6}=\frac{60}{12}
=5
$