7q-(q-4)=2q+4q(q+1)

Solution for 7q-(q-4)=2q+4q(q+1) equation:

7q-(q-4)=2q+4q(q+1)

We move all terms to the left:

7q-(q-4)-(2q+4q(q+1))=0

We get rid of parentheses

7q-q-(2q+4q(q+1))+4=0


We calculate terms in parentheses: -(2q+4q(q+1)), so:

2q+4q(q+1)

We multiply parentheses

4q^2+2q+4q

We add all the numbers together, and all the variables

4q^2+6q

Back to the equation:

-(4q^2+6q)


We add all the numbers together, and all the variables

6q-(4q^2+6q)+4=0

We get rid of parentheses

-4q^2+6q-6q+4=0

We add all the numbers together, and all the variables

-4q^2+4=0

a = -4; b = 0; c = +4;
Δ = b2-4ac
Δ = 02-4·(-4)·4
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8}{2*-4}=\frac{-8}{-8}
=1
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8}{2*-4}=\frac{8}{-8}
=-1
$