3/4p+6=p+10

Solution for 3/4p+6=p+10 equation:

3/4p+6=p+10

We move all terms to the left:

3/4p+6-(p+10)=0


Domain of the equation: 4p!=0

p!=0/4

p!=0

p∈R


We get rid of parentheses

3/4p-p-10+6=0

We multiply all the terms by the denominator


-p*4p-10*4p+6*4p+3=0

Wy multiply elements

-4p^2-40p+24p+3=0

We add all the numbers together, and all the variables

-4p^2-16p+3=0

a = -4; b = -16; c = +3;
Δ = b2-4ac
Δ = -162-4·(-4)·3
Δ = 304
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:

$\sqrt{\Delta}=\sqrt{304}=\sqrt{16*19}=\sqrt{16}*\sqrt{19}=4\sqrt{19}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-4\sqrt{19}}{2*-4}=\frac{16-4\sqrt{19}}{-8}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+4\sqrt{19}}{2*-4}=\frac{16+4\sqrt{19}}{-8}
$