2(p-1)-7(3p-2)=7p(p-4)

Solution for 2(p-1)-7(3p-2)=7p(p-4) equation:

2(p-1)-7(3p-2)=7p(p-4)

We move all terms to the left:

2(p-1)-7(3p-2)-(7p(p-4))=0

We multiply parentheses

2p-21p-(7p(p-4))-2+14=0


We calculate terms in parentheses: -(7p(p-4)), so:

7p(p-4)

We multiply parentheses

7p^2-28p

Back to the equation:

-(7p^2-28p)


We add all the numbers together, and all the variables

-19p-(7p^2-28p)+12=0

We get rid of parentheses

-7p^2-19p+28p+12=0

We add all the numbers together, and all the variables

-7p^2+9p+12=0

a = -7; b = 9; c = +12;
Δ = b2-4ac
Δ = 92-4·(-7)·12
Δ = 417
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{-(9)-\sqrt{417}}{2*-7}=\frac{-9-\sqrt{417}}{-14}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+\sqrt{417}}{2*-7}=\frac{-9+\sqrt{417}}{-14}
$