7y^2+18y+22=34y-41+6y^2

Solution for 7y^2+18y+22=34y-41+6y^2 equation:

7y^2+18y+22=34y-41+6y^2

We move all terms to the left:

7y^2+18y+22-(34y-41+6y^2)=0

We get rid of parentheses

7y^2-6y^2-34y+18y+41+22=0

We add all the numbers together, and all the variables

y^2-16y+63=0

a = 1; b = -16; c = +63;
Δ = b2-4ac
Δ = -162-4·1·63
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-2}{2*1}=\frac{14}{2}
=7
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+2}{2*1}=\frac{18}{2}
=9
$