-5(3-q)+4=1/5q-11

Solution for -5(3-q)+4=1/5q-11 equation:

-5(3-q)+4=1/5q-11

We move all terms to the left:

-5(3-q)+4-(1/5q-11)=0


Domain of the equation: 5q-11)!=0

q∈R


We add all the numbers together, and all the variables

-5(-1q+3)-(1/5q-11)+4=0

We multiply parentheses

5q-(1/5q-11)-15+4=0

We get rid of parentheses

5q-1/5q+11-15+4=0

We multiply all the terms by the denominator


5q*5q+11*5q-15*5q+4*5q-1=0

Wy multiply elements

25q^2+55q-75q+20q-1=0

We add all the numbers together, and all the variables

25q^2-1=0

a = 25; b = 0; c = -1;
Δ = b2-4ac
Δ = 02-4·25·(-1)
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-10}{2*25}=\frac{-10}{50}
=-1/5
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+10}{2*25}=\frac{10}{50}
=1/5
$