40=-100+(0.2q-3)q

Solution for 40=-100+(0.2q-3)q equation:

40=-100+(0.2q-3)q

We move all terms to the left:

40-(-100+(0.2q-3)q)=0


We calculate terms in parentheses: -(-100+(0.2q-3)q), so:

-100+(0.2q-3)q

determiningTheFunctionDomain
(0.2q-3)q-100

We multiply parentheses

0q^2-3q-100

We add all the numbers together, and all the variables

q^2-3q-100

Back to the equation:

-(q^2-3q-100)


We get rid of parentheses

-q^2+3q+100+40=0

We add all the numbers together, and all the variables

-1q^2+3q+140=0

a = -1; b = 3; c = +140;
Δ = b2-4ac
Δ = 32-4·(-1)·140
Δ = 569
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-\sqrt{569}}{2*-1}=\frac{-3-\sqrt{569}}{-2}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{569}}{2*-1}=\frac{-3+\sqrt{569}}{-2}
$