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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses

5q^2-15q-(5q-11)+4=0

We get rid of parentheses

5q^2-15q-5q+11+4=0

We add all the numbers together, and all the variables

5q^2-20q+15=0

a = 5; b = -20; c = +15;
Δ = b2-4ac
Δ = -202-4·5·15
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{-(-20)-10}{2*5}=\frac{10}{10}
=1
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+10}{2*5}=\frac{30}{10}
=3
$