4150=(1300-100p)p

Solution for 4150=(1300-100p)p equation:

4150=(1300-100p)p

We move all terms to the left:

4150-((1300-100p)p)=0

We add all the numbers together, and all the variables

-((-100p+1300)p)+4150=0


We calculate terms in parentheses: -((-100p+1300)p), so:

(-100p+1300)p

We multiply parentheses

-100p^2+1300p

Back to the equation:

-(-100p^2+1300p)


We get rid of parentheses

100p^2-1300p+4150=0

a = 100; b = -1300; c = +4150;
Δ = b2-4ac
Δ = -13002-4·100·4150
Δ = 30000
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:

$\sqrt{\Delta}=\sqrt{30000}=\sqrt{10000*3}=\sqrt{10000}*\sqrt{3}=100\sqrt{3}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1300)-100\sqrt{3}}{2*100}=\frac{1300-100\sqrt{3}}{200}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1300)+100\sqrt{3}}{2*100}=\frac{1300+100\sqrt{3}}{200}
$