Y=(x+5)(x+2)

Solution for Y=(x+5)(x+2) equation:

=(Y+5)(Y+2)

We move all terms to the left:

-((Y+5)(Y+2))=0

We multiply parentheses ..

-((+Y^2+2Y+5Y+10))=0


We calculate terms in parentheses: -((+Y^2+2Y+5Y+10)), so:

(+Y^2+2Y+5Y+10)

We get rid of parentheses

Y^2+2Y+5Y+10

We add all the numbers together, and all the variables

Y^2+7Y+10

Back to the equation:

-(Y^2+7Y+10)


We get rid of parentheses

-Y^2-7Y-10=0

We add all the numbers together, and all the variables

-1Y^2-7Y-10=0

a = -1; b = -7; c = -10;
Δ = b2-4ac
Δ = -72-4·(-1)·(-10)
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-3}{2*-1}=\frac{4}{-2}
=-2
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+3}{2*-1}=\frac{10}{-2}
=-5
$