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7k^2-6k-7=0
a = 7; b = -6; c = -7;
Δ = b2-4ac
Δ = -62-4·7·(-7)
Δ = 232
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}$$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{232}=\sqrt{4*58}=\sqrt{4}*\sqrt{58}=2\sqrt{58}$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{58}}{2*7}=\frac{6-2\sqrt{58}}{14} $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{58}}{2*7}=\frac{6+2\sqrt{58}}{14} $
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