(4y)2+24y+9=0

Solution for (4y)2+24y+9=0 equation:

(4y)2+24y+9=0

We add all the numbers together, and all the variables

4y^2+24y+9=0

a = 4; b = 24; c = +9;
Δ = b2-4ac
Δ = 242-4·4·9
Δ = 432
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{432}=\sqrt{144*3}=\sqrt{144}*\sqrt{3}=12\sqrt{3}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-12\sqrt{3}}{2*4}=\frac{-24-12\sqrt{3}}{8}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+12\sqrt{3}}{2*4}=\frac{-24+12\sqrt{3}}{8}
$