20(17+5p)=226+20p+p^2

Solution for 20(17+5p)=226+20p+p^2 equation:

20(17+5p)=226+20p+p^2

We move all terms to the left:

20(17+5p)-(226+20p+p^2)=0

We add all the numbers together, and all the variables

-(226+20p+p^2)+20(5p+17)=0

We multiply parentheses

-(226+20p+p^2)+100p+340=0

We get rid of parentheses

-p^2-20p+100p-226+340=0

We add all the numbers together, and all the variables

-1p^2+80p+114=0

a = -1; b = 80; c = +114;
Δ = b2-4ac
Δ = 802-4·(-1)·114
Δ = 6856
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:

$\sqrt{\Delta}=\sqrt{6856}=\sqrt{4*1714}=\sqrt{4}*\sqrt{1714}=2\sqrt{1714}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(80)-2\sqrt{1714}}{2*-1}=\frac{-80-2\sqrt{1714}}{-2}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(80)+2\sqrt{1714}}{2*-1}=\frac{-80+2\sqrt{1714}}{-2}
$