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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

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