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15w^2+6w-1=0
a = 15; b = 6; c = -1;
Δ = b2-4ac
Δ = 62-4·15·(-1)
Δ = 96
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{96}=\sqrt{16*6}=\sqrt{16}*\sqrt{6}=4\sqrt{6}$$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-4\sqrt{6}}{2*15}=\frac{-6-4\sqrt{6}}{30} $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+4\sqrt{6}}{2*15}=\frac{-6+4\sqrt{6}}{30} $
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