9y^2+10=53

Solution for 9y^2+10=53 equation:

9y^2+10=53

We move all terms to the left:

9y^2+10-(53)=0

We add all the numbers together, and all the variables

9y^2-43=0

a = 9; b = 0; c = -43;
Δ = b2-4ac
Δ = 02-4·9·(-43)
Δ = 1548
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:

$\sqrt{\Delta}=\sqrt{1548}=\sqrt{36*43}=\sqrt{36}*\sqrt{43}=6\sqrt{43}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{43}}{2*9}=\frac{0-6\sqrt{43}}{18}
=-\frac{6\sqrt{43}}{18}
=-\frac{\sqrt{43}}{3}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{43}}{2*9}=\frac{0+6\sqrt{43}}{18}
=\frac{6\sqrt{43}}{18}
=\frac{\sqrt{43}}{3}
$