10y+16=3y^2

Solution for 10y+16=3y^2 equation:

10y+16=3y^2

We move all terms to the left:

10y+16-(3y^2)=0

determiningTheFunctionDomain
-3y^2+10y+16=0

a = -3; b = 10; c = +16;
Δ = b2-4ac
Δ = 102-4·(-3)·16
Δ = 292
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:

$\sqrt{\Delta}=\sqrt{292}=\sqrt{4*73}=\sqrt{4}*\sqrt{73}=2\sqrt{73}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{73}}{2*-3}=\frac{-10-2\sqrt{73}}{-6}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{73}}{2*-3}=\frac{-10+2\sqrt{73}}{-6}
$