(-2p-10)(4p-2)=p-1-1

Solution for (-2p-10)(4p-2)=p-1-1 equation:

(-2p-10)(4p-2)=p-1-1

We move all terms to the left:

(-2p-10)(4p-2)-(p-1-1)=0

We add all the numbers together, and all the variables

(-2p-10)(4p-2)-(p-2)=0

We get rid of parentheses

(-2p-10)(4p-2)-p+2=0

We multiply parentheses ..

(-8p^2+4p-40p+20)-p+2=0

We add all the numbers together, and all the variables

(-8p^2+4p-40p+20)-1p+2=0

We get rid of parentheses

-8p^2+4p-40p-1p+20+2=0

We add all the numbers together, and all the variables

-8p^2-37p+22=0

a = -8; b = -37; c = +22;
Δ = b2-4ac
Δ = -372-4·(-8)·22
Δ = 2073
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-37)-\sqrt{2073}}{2*-8}=\frac{37-\sqrt{2073}}{-16}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-37)+\sqrt{2073}}{2*-8}=\frac{37+\sqrt{2073}}{-16}
$