P=-100+(0.7q-6)q

Solution for P=-100+(0.7q-6)q equation:

=-100+(0.7P-6)P

We move all terms to the left:

-(-100+(0.7P-6)P)=0


We calculate terms in parentheses: -(-100+(0.7P-6)P), so:

-100+(0.7P-6)P

determiningTheFunctionDomain
(0.7P-6)P-100

We multiply parentheses

0P^2-6P-100

We add all the numbers together, and all the variables

P^2-6P-100

Back to the equation:

-(P^2-6P-100)


We get rid of parentheses

-P^2+6P+100=0

We add all the numbers together, and all the variables

-1P^2+6P+100=0

a = -1; b = 6; c = +100;
Δ = b2-4ac
Δ = 62-4·(-1)·100
Δ = 436
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:

$\sqrt{\Delta}=\sqrt{436}=\sqrt{4*109}=\sqrt{4}*\sqrt{109}=2\sqrt{109}$
$P_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{109}}{2*-1}=\frac{-6-2\sqrt{109}}{-2}
$
$P_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{109}}{2*-1}=\frac{-6+2\sqrt{109}}{-2}
$