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5y^2-48=0
a = 5; b = 0; c = -48;
Δ = b2-4ac
Δ = 02-4·5·(-48)
Δ = 960
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{960}=\sqrt{64*15}=\sqrt{64}*\sqrt{15}=8\sqrt{15}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{15}}{2*5}=\frac{0-8\sqrt{15}}{10} =-\frac{8\sqrt{15}}{10} =-\frac{4\sqrt{15}}{5} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{15}}{2*5}=\frac{0+8\sqrt{15}}{10} =\frac{8\sqrt{15}}{10} =\frac{4\sqrt{15}}{5} $
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