A=150+(-05y)(y)

Solution for A=150+(-05y)(y) equation:

=150+(-05A)(A)

We move all terms to the left:

-(150+(-05A)(A))=0

We add all the numbers together, and all the variables

-(150+(-5A)A)=0


We calculate terms in parentheses: -(150+(-5A)A), so:

150+(-5A)A

determiningTheFunctionDomain
(-5A)A+150

We multiply parentheses

-5A^2+150

Back to the equation:

-(-5A^2+150)


We get rid of parentheses

5A^2-150=0

a = 5; b = 0; c = -150;
Δ = b2-4ac
Δ = 02-4·5·(-150)
Δ = 3000
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{3000}=\sqrt{100*30}=\sqrt{100}*\sqrt{30}=10\sqrt{30}$
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-10\sqrt{30}}{2*5}=\frac{0-10\sqrt{30}}{10}
=-\frac{10\sqrt{30}}{10}
=-\sqrt{30}
$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+10\sqrt{30}}{2*5}=\frac{0+10\sqrt{30}}{10}
=\frac{10\sqrt{30}}{10}
=\sqrt{30}
$