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15x^2-96=0
a = 15; b = 0; c = -96;
Δ = b2-4ac
Δ = 02-4·15·(-96)
Δ = 5760
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{5760}=\sqrt{576*10}=\sqrt{576}*\sqrt{10}=24\sqrt{10}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-24\sqrt{10}}{2*15}=\frac{0-24\sqrt{10}}{30} =-\frac{24\sqrt{10}}{30} =-\frac{4\sqrt{10}}{5} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+24\sqrt{10}}{2*15}=\frac{0+24\sqrt{10}}{30} =\frac{24\sqrt{10}}{30} =\frac{4\sqrt{10}}{5} $
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