4^3+24y=10y2

Solution for 4^3+24y=10y2 equation:

4^3+24y=10y^2

We move all terms to the left:

4^3+24y-(10y^2)=0

determiningTheFunctionDomain
-10y^2+24y+4^3=0

We add all the numbers together, and all the variables

-10y^2+24y+64=0

a = -10; b = 24; c = +64;
Δ = b2-4ac
Δ = 242-4·(-10)·64
Δ = 3136
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3136}=56$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-56}{2*-10}=\frac{-80}{-20}
=+4
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+56}{2*-10}=\frac{32}{-20}
=-1+3/5
$