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12y^2=100
We move all terms to the left:
12y^2-(100)=0
a = 12; b = 0; c = -100;
Δ = b2-4ac
Δ = 02-4·12·(-100)
Δ = 4800
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{4800}=\sqrt{1600*3}=\sqrt{1600}*\sqrt{3}=40\sqrt{3}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-40\sqrt{3}}{2*12}=\frac{0-40\sqrt{3}}{24} =-\frac{40\sqrt{3}}{24} =-\frac{5\sqrt{3}}{3} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+40\sqrt{3}}{2*12}=\frac{0+40\sqrt{3}}{24} =\frac{40\sqrt{3}}{24} =\frac{5\sqrt{3}}{3} $
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