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(2y)^2+y^2=10000
We move all terms to the left:
(2y)^2+y^2-(10000)=0
We add all the numbers together, and all the variables
3y^2-10000=0
a = 3; b = 0; c = -10000;
Δ = b2-4ac
Δ = 02-4·3·(-10000)
Δ = 120000
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{120000}=\sqrt{40000*3}=\sqrt{40000}*\sqrt{3}=200\sqrt{3}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-200\sqrt{3}}{2*3}=\frac{0-200\sqrt{3}}{6} =-\frac{200\sqrt{3}}{6} =-\frac{100\sqrt{3}}{3} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+200\sqrt{3}}{2*3}=\frac{0+200\sqrt{3}}{6} =\frac{200\sqrt{3}}{6} =\frac{100\sqrt{3}}{3} $
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