7q-(q-3)=3q+3q(q+1)

Solution for 7q-(q-3)=3q+3q(q+1) equation:

7q-(q-3)=3q+3q(q+1)

We move all terms to the left:

7q-(q-3)-(3q+3q(q+1))=0

We get rid of parentheses

7q-q-(3q+3q(q+1))+3=0


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

3q+3q(q+1)

We multiply parentheses

3q^2+3q+3q

We add all the numbers together, and all the variables

3q^2+6q

Back to the equation:

-(3q^2+6q)


We add all the numbers together, and all the variables

6q-(3q^2+6q)+3=0

We get rid of parentheses

-3q^2+6q-6q+3=0

We add all the numbers together, and all the variables

-3q^2+3=0

a = -3; b = 0; c = +3;
Δ = b2-4ac
Δ = 02-4·(-3)·3
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6}{2*-3}=\frac{-6}{-6}
=1
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6}{2*-3}=\frac{6}{-6}
=-1
$