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b^2=100-2b^2
We move all terms to the left:
b^2-(100-2b^2)=0
We get rid of parentheses
b^2+2b^2-100=0
We add all the numbers together, and all the variables
3b^2-100=0
a = 3; b = 0; c = -100;
Δ = b2-4ac
Δ = 02-4·3·(-100)
Δ = 1200
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1200}=\sqrt{400*3}=\sqrt{400}*\sqrt{3}=20\sqrt{3}$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-20\sqrt{3}}{2*3}=\frac{0-20\sqrt{3}}{6} =-\frac{20\sqrt{3}}{6} =-\frac{10\sqrt{3}}{3} $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+20\sqrt{3}}{2*3}=\frac{0+20\sqrt{3}}{6} =\frac{20\sqrt{3}}{6} =\frac{10\sqrt{3}}{3} $
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