2P+100=5(8x^2-20x)+1

Solution for 2P+100=5(8x^2-20x)+1 equation:

2+100=5(8P^2-20P)+1

We move all terms to the left:

2+100-(5(8P^2-20P)+1)=0

We add all the numbers together, and all the variables

-(5(8P^2-20P)+1)+102=0


We calculate terms in parentheses: -(5(8P^2-20P)+1), so:

5(8P^2-20P)+1

We multiply parentheses

40P^2-100P+1

Back to the equation:

-(40P^2-100P+1)


We get rid of parentheses

-40P^2+100P-1+102=0

We add all the numbers together, and all the variables

-40P^2+100P+101=0

a = -40; b = 100; c = +101;
Δ = b2-4ac
Δ = 1002-4·(-40)·101
Δ = 26160
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:

$\sqrt{\Delta}=\sqrt{26160}=\sqrt{16*1635}=\sqrt{16}*\sqrt{1635}=4\sqrt{1635}$
$P_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(100)-4\sqrt{1635}}{2*-40}=\frac{-100-4\sqrt{1635}}{-80}
$
$P_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(100)+4\sqrt{1635}}{2*-40}=\frac{-100+4\sqrt{1635}}{-80}
$