P=-100+(0.7q-10)q

Solution for P=-100+(0.7q-10)q equation:

=-100+(0.7P-10)P

We move all terms to the left:

-(-100+(0.7P-10)P)=0


We calculate terms in parentheses: -(-100+(0.7P-10)P), so:

-100+(0.7P-10)P

determiningTheFunctionDomain
(0.7P-10)P-100

We multiply parentheses

0P^2-10P-100

We add all the numbers together, and all the variables

P^2-10P-100

Back to the equation:

-(P^2-10P-100)


We get rid of parentheses

-P^2+10P+100=0

We add all the numbers together, and all the variables

-1P^2+10P+100=0

a = -1; b = 10; c = +100;
Δ = b2-4ac
Δ = 102-4·(-1)·100
Δ = 500
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$P_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$P_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{500}=\sqrt{100*5}=\sqrt{100}*\sqrt{5}=10\sqrt{5}$
$P_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-10\sqrt{5}}{2*-1}=\frac{-10-10\sqrt{5}}{-2}
$
$P_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+10\sqrt{5}}{2*-1}=\frac{-10+10\sqrt{5}}{-2}
$