(1/6)(12p+24)=6

Solution for (1/6)(12p+24)=6 equation:

(1/6)(12p+24)=6

We move all terms to the left:

(1/6)(12p+24)-(6)=0


Domain of the equation: 6)(12p+24)!=0

p∈R


We add all the numbers together, and all the variables

(+1/6)(12p+24)-6=0

We multiply parentheses ..

(+12p^2+1/6*24)-6=0

We multiply all the terms by the denominator


(+12p^2+1-6*6*24)=0

We get rid of parentheses

12p^2+1-6*6*24=0

We add all the numbers together, and all the variables

12p^2-863=0

a = 12; b = 0; c = -863;
Δ = b2-4ac
Δ = 02-4·12·(-863)
Δ = 41424
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:

$\sqrt{\Delta}=\sqrt{41424}=\sqrt{16*2589}=\sqrt{16}*\sqrt{2589}=4\sqrt{2589}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{2589}}{2*12}=\frac{0-4\sqrt{2589}}{24}
=-\frac{4\sqrt{2589}}{24}
=-\frac{\sqrt{2589}}{6}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{2589}}{2*12}=\frac{0+4\sqrt{2589}}{24}
=\frac{4\sqrt{2589}}{24}
=\frac{\sqrt{2589}}{6}
$