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101y^{2}-243=98y^{2}
Combine 111y^{2} and -10y^{2} to get 101y^{2}.
101y^{2}-243-98y^{2}=0
Subtract 98y^{2} from both sides.
3y^{2}-243=0
Combine 101y^{2} and -98y^{2} to get 3y^{2}.
y^{2}-81=0
Divide both sides by 3.
\left(y-9\right)\left(y+9\right)=0
Consider y^{2}-81. Rewrite y^{2}-81 as y^{2}-9^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).
y=9 y=-9
To find equation solutions, solve y-9=0 and y+9=0.
101y^{2}-243=98y^{2}
Combine 111y^{2} and -10y^{2} to get 101y^{2}.
101y^{2}-243-98y^{2}=0
Subtract 98y^{2} from both sides.
3y^{2}-243=0
Combine 101y^{2} and -98y^{2} to get 3y^{2}.
3y^{2}=243
Add 243 to both sides. Anything plus zero gives itself.
y^{2}=\frac{243}{3}
Divide both sides by 3.
y^{2}=81
Divide 243 by 3 to get 81.
y=9 y=-9
Take the square root of both sides of the equation.
101y^{2}-243=98y^{2}
Combine 111y^{2} and -10y^{2} to get 101y^{2}.
101y^{2}-243-98y^{2}=0
Subtract 98y^{2} from both sides.
3y^{2}-243=0
Combine 101y^{2} and -98y^{2} to get 3y^{2}.
y=\frac{0±\sqrt{0^{2}-4\times 3\left(-243\right)}}{2\times 3}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 0 for b, and -243 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
y=\frac{0±\sqrt{-4\times 3\left(-243\right)}}{2\times 3}
Square 0.
y=\frac{0±\sqrt{-12\left(-243\right)}}{2\times 3}
Multiply -4 times 3.
y=\frac{0±\sqrt{2916}}{2\times 3}
Multiply -12 times -243.
y=\frac{0±54}{2\times 3}
Take the square root of 2916.
y=\frac{0±54}{6}
Multiply 2 times 3.
y=9
Now solve the equation y=\frac{0±54}{6} when ± is plus. Divide 54 by 6.
y=-9
Now solve the equation y=\frac{0±54}{6} when ± is minus. Divide -54 by 6.
y=9 y=-9
The equation is now solved.