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4k^2+17k-246=0
a = 4; b = 17; c = -246;
Δ = b2-4ac
Δ = 172-4·4·(-246)
Δ = 4225
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{4225}=65$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-65}{2*4}=\frac{-82}{8} =-10+1/4 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+65}{2*4}=\frac{48}{8} =6 $
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