y^2+14y+38=3y+8

Solution for y^2+14y+38=3y+8 equation:

y^2+14y+38=3y+8

We move all terms to the left:

y^2+14y+38-(3y+8)=0

We get rid of parentheses

y^2+14y-3y-8+38=0

We add all the numbers together, and all the variables

y^2+11y+30=0

a = 1; b = 11; c = +30;
Δ = b2-4ac
Δ = 112-4·1·30
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-1}{2*1}=\frac{-12}{2}
=-6
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+1}{2*1}=\frac{-10}{2}
=-5
$