Y=-(x+2)(x+2)+9

Solution for Y=-(x+2)(x+2)+9 equation:

=-(Y+2)(Y+2)+9

We move all terms to the left:

-(-(Y+2)(Y+2)+9)=0

We multiply parentheses ..

-(-(+Y^2+2Y+2Y+4)+9)=0


We calculate terms in parentheses: -(-(+Y^2+2Y+2Y+4)+9), so:

-(+Y^2+2Y+2Y+4)+9

We get rid of parentheses

-Y^2-2Y-2Y-4+9

We add all the numbers together, and all the variables

-1Y^2-4Y+5

Back to the equation:

-(-1Y^2-4Y+5)


We get rid of parentheses

1Y^2+4Y-5=0

We add all the numbers together, and all the variables

Y^2+4Y-5=0

a = 1; b = 4; c = -5;
Δ = b2-4ac
Δ = 42-4·1·(-5)
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-6}{2*1}=\frac{-10}{2}
=-5
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+6}{2*1}=\frac{2}{2}
=1
$