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5y^2-15=0
a = 5; b = 0; c = -15;
Δ = b2-4ac
Δ = 02-4·5·(-15)
Δ = 300
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{300}=\sqrt{100*3}=\sqrt{100}*\sqrt{3}=10\sqrt{3}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-10\sqrt{3}}{2*5}=\frac{0-10\sqrt{3}}{10} =-\frac{10\sqrt{3}}{10} =-\sqrt{3} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+10\sqrt{3}}{2*5}=\frac{0+10\sqrt{3}}{10} =\frac{10\sqrt{3}}{10} =\sqrt{3} $
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