3(x^2-7x-1)+5=2(x+1)

Solution for 3(x^2-7x-1)+5=2(x+1) equation:

3(x^2-7x-1)+5=2(x+1)

We move all terms to the left:

3(x^2-7x-1)+5-(2(x+1))=0

We multiply parentheses

3x^2-21x-(2(x+1))-3+5=0


We calculate terms in parentheses: -(2(x+1)), so:

2(x+1)

We multiply parentheses

2x+2

Back to the equation:

-(2x+2)


We add all the numbers together, and all the variables

3x^2-21x-(2x+2)+2=0

We get rid of parentheses

3x^2-21x-2x-2+2=0

We add all the numbers together, and all the variables

3x^2-23x=0

a = 3; b = -23; c = 0;
Δ = b2-4ac
Δ = -232-4·3·0
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-23)-23}{2*3}=\frac{0}{6}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+23}{2*3}=\frac{46}{6}
=7+2/3
$