0=9y^2+42y+47

Solution for 0=9y^2+42y+47 equation:

0=9y^2+42y+47

We move all terms to the left:

0-(9y^2+42y+47)=0

We add all the numbers together, and all the variables

-(9y^2+42y+47)=0

We get rid of parentheses

-9y^2-42y-47=0

a = -9; b = -42; c = -47;
Δ = b2-4ac
Δ = -422-4·(-9)·(-47)
Δ = 72
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:

$\sqrt{\Delta}=\sqrt{72}=\sqrt{36*2}=\sqrt{36}*\sqrt{2}=6\sqrt{2}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-42)-6\sqrt{2}}{2*-9}=\frac{42-6\sqrt{2}}{-18}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-42)+6\sqrt{2}}{2*-9}=\frac{42+6\sqrt{2}}{-18}
$