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19k^2+6k-64=0
a = 19; b = 6; c = -64;
Δ = b2-4ac
Δ = 62-4·19·(-64)
Δ = 4900
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}$$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{4900}=70$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-70}{2*19}=\frac{-76}{38} =-2 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+70}{2*19}=\frac{64}{38} =1+13/19 $
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