10p(2p+5)=p2+

Solution for 10p(2p+5)=p2+ equation:

10p(2p+5)=p2+

We move all terms to the left:

10p(2p+5)-(p2+)=0

We add all the numbers together, and all the variables

-(+p^2+)+10p(2p+5)=0

We multiply parentheses

-(+p^2+)+20p^2+50p=0

We get rid of parentheses

-p^2+20p^2+50p-=0

We add all the numbers together, and all the variables

19p^2+50p=0

a = 19; b = 50; c = 0;
Δ = b2-4ac
Δ = 502-4·19·0
Δ = 2500
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2500}=50$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(50)-50}{2*19}=\frac{-100}{38}
=-2+12/19
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(50)+50}{2*19}=\frac{0}{38}
=0
$