(3/5)(p+2)=2

Solution for (3/5)(p+2)=2 equation:

(3/5)(p+2)=2

We move all terms to the left:

(3/5)(p+2)-(2)=0


Domain of the equation: 5)(p+2)!=0

p∈R


We add all the numbers together, and all the variables

(+3/5)(p+2)-2=0

We multiply parentheses ..

(+3p^2+3/5*2)-2=0

We multiply all the terms by the denominator


(+3p^2+3-2*5*2)=0

We get rid of parentheses

3p^2+3-2*5*2=0

We add all the numbers together, and all the variables

3p^2-17=0

a = 3; b = 0; c = -17;
Δ = b2-4ac
Δ = 02-4·3·(-17)
Δ = 204
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:

$\sqrt{\Delta}=\sqrt{204}=\sqrt{4*51}=\sqrt{4}*\sqrt{51}=2\sqrt{51}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{51}}{2*3}=\frac{0-2\sqrt{51}}{6}
=-\frac{2\sqrt{51}}{6}
=-\frac{\sqrt{51}}{3}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{51}}{2*3}=\frac{0+2\sqrt{51}}{6}
=\frac{2\sqrt{51}}{6}
=\frac{\sqrt{51}}{3}
$