A=-l^2+10l

Solution for A=-l^2+10l equation:

=-A^2+10A

We move all terms to the left:

-(-A^2+10A)=0

We get rid of parentheses

A^2-10A=0

a = 1; b = -10; c = 0;
Δ = b2-4ac
Δ = -102-4·1·0
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-10}{2*1}=\frac{0}{2}
=0
$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+10}{2*1}=\frac{20}{2}
=10
$