Q=50q-(10+25q+q^2)

Solution for Q=50q-(10+25q+q^2) equation:

=50Q-(10+25Q+Q^2)

We move all terms to the left:

-(50Q-(10+25Q+Q^2))=0


We calculate terms in parentheses: -(50Q-(10+25Q+Q^2)), so:

50Q-(10+25Q+Q^2)

determiningTheFunctionDomain
-(10+25Q+Q^2)+50Q

We get rid of parentheses

-Q^2-25Q+50Q-10

We add all the numbers together, and all the variables

-1Q^2+25Q-10

Back to the equation:

-(-1Q^2+25Q-10)


We get rid of parentheses

1Q^2-25Q+10=0

We add all the numbers together, and all the variables

Q^2-25Q+10=0

a = 1; b = -25; c = +10;
Δ = b2-4ac
Δ = -252-4·1·10
Δ = 585
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{585}=\sqrt{9*65}=\sqrt{9}*\sqrt{65}=3\sqrt{65}$
$Q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-25)-3\sqrt{65}}{2*1}=\frac{25-3\sqrt{65}}{2}
$
$Q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+3\sqrt{65}}{2*1}=\frac{25+3\sqrt{65}}{2}
$