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

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

=(2A-3)(4A-7)

We move all terms to the left:

-((2A-3)(4A-7))=0

We multiply parentheses ..

-((+8A^2-14A-12A+21))=0


We calculate terms in parentheses: -((+8A^2-14A-12A+21)), so:

(+8A^2-14A-12A+21)

We get rid of parentheses

8A^2-14A-12A+21

We add all the numbers together, and all the variables

8A^2-26A+21

Back to the equation:

-(8A^2-26A+21)


We get rid of parentheses

-8A^2+26A-21=0

a = -8; b = 26; c = -21;
Δ = b2-4ac
Δ = 262-4·(-8)·(-21)
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(26)-2}{2*-8}=\frac{-28}{-16}
=1+3/4
$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(26)+2}{2*-8}=\frac{-24}{-16}
=1+1/2
$