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

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

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

We move all terms to the left:

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

We multiply parentheses ..

(+8p^2+6p-28p-21)-3=0

We get rid of parentheses

8p^2+6p-28p-21-3=0

We add all the numbers together, and all the variables

8p^2-22p-24=0

a = 8; b = -22; c = -24;
Δ = b2-4ac
Δ = -222-4·8·(-24)
Δ = 1252
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:

$\sqrt{\Delta}=\sqrt{1252}=\sqrt{4*313}=\sqrt{4}*\sqrt{313}=2\sqrt{313}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-2\sqrt{313}}{2*8}=\frac{22-2\sqrt{313}}{16}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+2\sqrt{313}}{2*8}=\frac{22+2\sqrt{313}}{16}
$