(6p2+4p+5)=(2p2-5p+1)

Solution for (6p2+4p+5)=(2p2-5p+1) equation:

(6p^2+4p+5)=(2p^2-5p+1)

We move all terms to the left:

(6p^2+4p+5)-((2p^2-5p+1))=0

We get rid of parentheses

6p^2+4p-((2p^2-5p+1))+5=0


We calculate terms in parentheses: -((2p^2-5p+1)), so:

(2p^2-5p+1)

We get rid of parentheses

2p^2-5p+1

Back to the equation:

-(2p^2-5p+1)


We get rid of parentheses

6p^2-2p^2+4p+5p-1+5=0

We add all the numbers together, and all the variables

4p^2+9p+4=0

a = 4; b = 9; c = +4;
Δ = b2-4ac
Δ = 92-4·4·4
Δ = 17
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-\sqrt{17}}{2*4}=\frac{-9-\sqrt{17}}{8}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+\sqrt{17}}{2*4}=\frac{-9+\sqrt{17}}{8}
$