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 get rid of 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

32p^2-2=0

a = 32; b = 0; c = -2;
Δ = b2-4ac
Δ = 02-4·32·(-2)
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16}{2*32}=\frac{-16}{64}
=-1/4
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16}{2*32}=\frac{16}{64}
=1/4
$