Question 1203435
<font color=black size=3>
The tutor @ikleyn has shown two great approaches.


I'll show two more methods.


Instead of y, I'll use x.


The task is to solve x^2 = 81 which is the same as x^2-81 = 0


Further we can write 
1x^2 + 0x + (-81) = 0


Compare that to the template
ax^2 + bx + c = 0
to find that:
a = 1
b = 0
c = -81


Those values are plugged into the quadratic formula.
{{{x = (-b+-sqrt(b^2-4ac))/(2a)}}}


{{{x = (-0+-sqrt((0)^2-4(1)(-81)))/(2(1))}}}


{{{x = (0+-sqrt(324))/(2)}}}


{{{x = (0+-  18)/(2)}}}


{{{x = (0+18)/(2)}}} or {{{x = (0-18)/(2)}}}


{{{x = (18)/(2)}}} or  {{{x = (-18)/(2)}}}


{{{x = 9}}} or  {{{x = -9}}}
While this method takes a bit more work, I think it's still handy to know multiple approaches. 


----------------------------------------


Another approach is to graph y = x^2 - 81 using a tool like Desmos or GeoGebra.
This produces a parabola that opens upward. The parabola crosses the x axis at x = 9 and x = -9, which are the two solutions to x^2 = 81.


{{{graph(400,400,-10,10,-100,100,-700,x^2-81)}}}
Window: xmin = -10, xmax = 10, ymin = -100, ymax = 100


Link to the interactive Desmos graph
<a href="https://www.desmos.com/calculator/ewdbmqwf2c">https://www.desmos.com/calculator/ewdbmqwf2c</a>


Therefore, the two solutions to y^2 = 81 are y = 9 or y = -9


As a quick check,
y^2 = 81
9^2 = 81
9*9 = 81
81 = 81
and also
y^2 = 81
(-9)^2 = 81
(-9)*(-9) = 81
81 = 81
When squaring the negative, the two negatives cancel out to get a positive.
In other words: negative times negative = positive
We get the same thing on both sides (81), so we have confirmed the answers are correct.
</font>