3p^2-42p+147=3

Solution for 3p^2-42p+147=3 equation:

3p^2-42p+147=3

We move all terms to the left:

3p^2-42p+147-(3)=0

We add all the numbers together, and all the variables

3p^2-42p+144=0

a = 3; b = -42; c = +144;
Δ = b2-4ac
Δ = -422-4·3·144
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-42)-6}{2*3}=\frac{36}{6}
=6
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-42)+6}{2*3}=\frac{48}{6}
=8
$