7*(9^2p-4)+3=45

Solution for 7*(9^2p-4)+3=45 equation:

7(9^2p-4)+3=45

We move all terms to the left:

7(9^2p-4)+3-(45)=0

We add all the numbers together, and all the variables

7(9^2p-4)-42=0

We multiply parentheses

63p^2-28-42=0

We add all the numbers together, and all the variables

63p^2-70=0

a = 63; b = 0; c = -70;
Δ = b2-4ac
Δ = 02-4·63·(-70)
Δ = 17640
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:

$\sqrt{\Delta}=\sqrt{17640}=\sqrt{1764*10}=\sqrt{1764}*\sqrt{10}=42\sqrt{10}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-42\sqrt{10}}{2*63}=\frac{0-42\sqrt{10}}{126}
=-\frac{42\sqrt{10}}{126}
=-\frac{\sqrt{10}}{3}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+42\sqrt{10}}{2*63}=\frac{0+42\sqrt{10}}{126}
=\frac{42\sqrt{10}}{126}
=\frac{\sqrt{10}}{3}
$