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7k^2-35=0
a = 7; b = 0; c = -35;
Δ = b2-4ac
Δ = 02-4·7·(-35)
Δ = 980
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}$$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{980}=\sqrt{196*5}=\sqrt{196}*\sqrt{5}=14\sqrt{5}$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-14\sqrt{5}}{2*7}=\frac{0-14\sqrt{5}}{14} =-\frac{14\sqrt{5}}{14} =-\sqrt{5} $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+14\sqrt{5}}{2*7}=\frac{0+14\sqrt{5}}{14} =\frac{14\sqrt{5}}{14} =\sqrt{5} $
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