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

Solution for Y=(r+5)(r-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+3Y-10

Back to the equation:

-(Y^2+3Y-10)


We get rid of parentheses

-Y^2-3Y+10=0

We add all the numbers together, and all the variables

-1Y^2-3Y+10=0

a = -1; b = -3; c = +10;
Δ = b2-4ac
Δ = -32-4·(-1)·10
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-7}{2*-1}=\frac{-4}{-2}
=+2
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+7}{2*-1}=\frac{10}{-2}
=-5
$