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15x^2-5=0
a = 15; b = 0; c = -5;
Δ = b2-4ac
Δ = 02-4·15·(-5)
Δ = 300
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{300}=\sqrt{100*3}=\sqrt{100}*\sqrt{3}=10\sqrt{3}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-10\sqrt{3}}{2*15}=\frac{0-10\sqrt{3}}{30} =-\frac{10\sqrt{3}}{30} =-\frac{\sqrt{3}}{3} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+10\sqrt{3}}{2*15}=\frac{0+10\sqrt{3}}{30} =\frac{10\sqrt{3}}{30} =\frac{\sqrt{3}}{3} $
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