-(p+4)+40p2=

Solution for -(p+4)+40p2= equation:

-(p+4)+40p^2=

We move all terms to the left:

-(p+4)+40p^2-()=0

We add all the numbers together, and all the variables

40p^2-(p+4)=0

We get rid of parentheses

40p^2-p-4=0

We add all the numbers together, and all the variables

40p^2-1p-4=0

a = 40; b = -1; c = -4;
Δ = b2-4ac
Δ = -12-4·40·(-4)
Δ = 641
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{-(-1)-\sqrt{641}}{2*40}=\frac{1-\sqrt{641}}{80}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{641}}{2*40}=\frac{1+\sqrt{641}}{80}
$