(3x^2-11x-4)=(3x+1)

Solution for (3x^2-11x-4)=(3x+1) equation:

(3x^2-11x-4)=(3x+1)

We move all terms to the left:

(3x^2-11x-4)-((3x+1))=0

We get rid of parentheses

3x^2-11x-((3x+1))-4=0


We calculate terms in parentheses: -((3x+1)), so:

(3x+1)

We get rid of parentheses

3x+1

Back to the equation:

-(3x+1)


We get rid of parentheses

3x^2-11x-3x-1-4=0

We add all the numbers together, and all the variables

3x^2-14x-5=0

a = 3; b = -14; c = -5;
Δ = b2-4ac
Δ = -142-4·3·(-5)
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{-(-14)-16}{2*3}=\frac{-2}{6}
=-1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+16}{2*3}=\frac{30}{6}
=5
$