(y^2+4)-6(7+y)=-22

Solution for (y^2+4)-6(7+y)=-22 equation:

(y^2+4)-6(7+y)=-22

We move all terms to the left:

(y^2+4)-6(7+y)-(-22)=0

We add all the numbers together, and all the variables

(y^2+4)-6(y+7)-(-22)=0

We add all the numbers together, and all the variables

(y^2+4)-6(y+7)+22=0

We multiply parentheses

(y^2+4)-6y-42+22=0

We get rid of parentheses

y^2-6y+4-42+22=0

We add all the numbers together, and all the variables

y^2-6y-16=0

a = 1; b = -6; c = -16;
Δ = b2-4ac
Δ = -62-4·1·(-16)
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-10}{2*1}=\frac{-4}{2}
=-2
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+10}{2*1}=\frac{16}{2}
=8
$