40p^2=24-28p

Solution for 40p^2=24-28p equation:

40p^2=24-28p

We move all terms to the left:

40p^2-(24-28p)=0

We add all the numbers together, and all the variables

40p^2-(-28p+24)=0

We get rid of parentheses

40p^2+28p-24=0

a = 40; b = 28; c = -24;
Δ = b2-4ac
Δ = 282-4·40·(-24)
Δ = 4624
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4624}=68$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-68}{2*40}=\frac{-96}{80}
=-1+1/5
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+68}{2*40}=\frac{40}{80}
=1/2
$