2y^2+18y+41=(y+6)

Solution for 2y^2+18y+41=(y+6) equation:

2y^2+18y+41=(y+6)

We move all terms to the left:

2y^2+18y+41-((y+6))=0


We calculate terms in parentheses: -((y+6)), so:

(y+6)

We get rid of parentheses

y+6

Back to the equation:

-(y+6)


We get rid of parentheses

2y^2+18y-y-6+41=0

We add all the numbers together, and all the variables

2y^2+17y+35=0

a = 2; b = 17; c = +35;
Δ = b2-4ac
Δ = 172-4·2·35
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{-(17)-3}{2*2}=\frac{-20}{4}
=-5
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+3}{2*2}=\frac{-14}{4}
=-3+1/2
$