(y^2-18-y)+(10=4y-5y^2)

Solution for (y^2-18-y)+(10=4y-5y^2) equation:

(y^2-18-y)+(10=4y-5y^2)

We move all terms to the left:

(y^2-18-y)+(10-(4y-5y^2))=0

We get rid of parentheses

(10-(4y-5y^2))+y^2-y-18=0


We calculate terms in parentheses: +(10-(4y-5y^2)), so:

10-(4y-5y^2)

determiningTheFunctionDomain
-(4y-5y^2)+10

We get rid of parentheses

5y^2-4y+10

Back to the equation:

+(5y^2-4y+10)


We add all the numbers together, and all the variables

y^2-1y+(5y^2-4y+10)-18=0

We get rid of parentheses

y^2+5y^2-1y-4y+10-18=0

We add all the numbers together, and all the variables

6y^2-5y-8=0

a = 6; b = -5; c = -8;
Δ = b2-4ac
Δ = -52-4·6·(-8)
Δ = 217
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-\sqrt{217}}{2*6}=\frac{5-\sqrt{217}}{12}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{217}}{2*6}=\frac{5+\sqrt{217}}{12}
$