(4y)2=y+4

Solution for (4y)2=y+4 equation:

(4y)2=y+4

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

4y^2-y-4=0

We add all the numbers together, and all the variables

4y^2-1y-4=0

a = 4; b = -1; c = -4;
Δ = b2-4ac
Δ = -12-4·4·(-4)
Δ = 65
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-\sqrt{65}}{2*4}=\frac{1-\sqrt{65}}{8}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{65}}{2*4}=\frac{1+\sqrt{65}}{8}
$