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15k^2-26k+3=0
a = 15; b = -26; c = +3;
Δ = b2-4ac
Δ = -262-4·15·3
Δ = 496
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{496}=\sqrt{16*31}=\sqrt{16}*\sqrt{31}=4\sqrt{31}$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-26)-4\sqrt{31}}{2*15}=\frac{26-4\sqrt{31}}{30} $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-26)+4\sqrt{31}}{2*15}=\frac{26+4\sqrt{31}}{30} $
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