Is this a quadratic equation?

Question

I have this equation:

$-19y^2 + 5y\sqrt{y} + 10y + 12 = 0$

I'm stuck with finding $y$. I tried numerous adjustments like this for example:

$y*(-19y+5\sqrt{y}+10)+12 =0$

but it didn't get me any closer to a solution. I know it must be really simple but right now I'm stuck for hours. Can you give me a hint how to solve this equation?

Answer 1

Is this a quadratic equation?

No, because for that to be the case, all exponents would have to be natural numbers. But $\sqrt y=y^\frac12$ and $\frac12\not\in$ N. However, by letting $x=\sqrt y$, you'll have a quartic equation in x.

Answer 2

$−19y2+5y√y+10y+12=0$

Assuming

$\sqrt y=t$ then, $y=t^2,y^2=t^4$

So, it becomes $-19t^4+5t^3+10t^2+12=0$

You see that it is missing expression with $t$, so that will make this difficult to calculate. Alternatively, I would suggest throwing the whole expression on a graph and find the x-intersects.

you will get a value of $t\approx 1.14$ , so your $y$ will become 1.299