4p2+-9p+11=p2+-5p+4

Solution for 4p2+-9p+11=p2+-5p+4 equation:

4p^2+-9p+11=p2+-5p+4

We move all terms to the left:

4p^2+-9p+11-(p2+-5p+4)=0

We add all the numbers together, and all the variables

4p^2-(+p^2-5p++4)-9p+11+=0

We add all the numbers together, and all the variables

4p^2-(+p^2-5p++4)-9p=0

We use the square of the difference formula

4p^2-(+p^2-5p+4)-9p=0

We get rid of parentheses

4p^2-p^2+5p-9p-4=0

We add all the numbers together, and all the variables

3p^2-4p-4=0

a = 3; b = -4; c = -4;
Δ = b2-4ac
Δ = -42-4·3·(-4)
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-8}{2*3}=\frac{-4}{6}
=-2/3
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+8}{2*3}=\frac{12}{6}
=2
$