10=p*(4-2)

Solution for 10=p*(4-2) equation:

10=p(4-2)

We move all terms to the left:

10-(p(4-2))=0

We add all the numbers together, and all the variables

-(p2)+10=0

We add all the numbers together, and all the variables

-1p^2+10=0

a = -1; b = 0; c = +10;
Δ = b2-4ac
Δ = 02-4·(-1)·10
Δ = 40
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:

$\sqrt{\Delta}=\sqrt{40}=\sqrt{4*10}=\sqrt{4}*\sqrt{10}=2\sqrt{10}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{10}}{2*-1}=\frac{0-2\sqrt{10}}{-2}
=-\frac{2\sqrt{10}}{-2}
=-\frac{\sqrt{10}}{-1}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{10}}{2*-1}=\frac{0+2\sqrt{10}}{-2}
=\frac{2\sqrt{10}}{-2}
=\frac{\sqrt{10}}{-1}
$