21+4y=y^2

Solution for 21+4y=y^2 equation:

21+4y=y^2

We move all terms to the left:

21+4y-(y^2)=0

determiningTheFunctionDomain
-y^2+4y+21=0

We add all the numbers together, and all the variables

-1y^2+4y+21=0

a = -1; b = 4; c = +21;
Δ = b2-4ac
Δ = 42-4·(-1)·21
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{-(4)-10}{2*-1}=\frac{-14}{-2}
=+7
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+10}{2*-1}=\frac{6}{-2}
=-3
$