p^2-(7/4)*p+2/4=0

Solution for p^2-(7/4)*p+2/4=0 equation:

p^2-(7/4)*p+2/4=0


Domain of the equation: 4)*p!=0

p!=0/1

p!=0

p∈R


We add all the numbers together, and all the variables

p^2-(+7/4)*p+2/4=0

We multiply parentheses

p^2-7p^2+2/4=0

We multiply all the terms by the denominator


p^2*4-7p^2*4+2=0

Wy multiply elements

4p^2-28p^2+2=0

We add all the numbers together, and all the variables

-24p^2+2=0

a = -24; b = 0; c = +2;
Δ = b2-4ac
Δ = 02-4·(-24)·2
Δ = 192
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:

$\sqrt{\Delta}=\sqrt{192}=\sqrt{64*3}=\sqrt{64}*\sqrt{3}=8\sqrt{3}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{3}}{2*-24}=\frac{0-8\sqrt{3}}{-48}
=-\frac{8\sqrt{3}}{-48}
=-\frac{\sqrt{3}}{-6}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{3}}{2*-24}=\frac{0+8\sqrt{3}}{-48}
=\frac{8\sqrt{3}}{-48}
=\frac{\sqrt{3}}{-6}
$