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7y^2=12
We move all terms to the left:
7y^2-(12)=0
a = 7; b = 0; c = -12;
Δ = b2-4ac
Δ = 02-4·7·(-12)
Δ = 336
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{336}=\sqrt{16*21}=\sqrt{16}*\sqrt{21}=4\sqrt{21}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{21}}{2*7}=\frac{0-4\sqrt{21}}{14} =-\frac{4\sqrt{21}}{14} =-\frac{2\sqrt{21}}{7} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{21}}{2*7}=\frac{0+4\sqrt{21}}{14} =\frac{4\sqrt{21}}{14} =\frac{2\sqrt{21}}{7} $
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