4y+(y*y)=96

Solution for 4y+(y*y)=96 equation:

4y+(y*y)=96

We move all terms to the left:

4y+(y*y)-(96)=0

We add all the numbers together, and all the variables

4y+(+y*y)-96=0

We get rid of parentheses

4y+y*y-96=0

Wy multiply elements

y^2+4y-96=0

a = 1; b = 4; c = -96;
Δ = b2-4ac
Δ = 42-4·1·(-96)
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-20}{2*1}=\frac{-24}{2}
=-12
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+20}{2*1}=\frac{16}{2}
=8
$