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6r^2+2r-1=0
a = 6; b = 2; c = -1;
Δ = b2-4ac
Δ = 22-4·6·(-1)
Δ = 28
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{28}=\sqrt{4*7}=\sqrt{4}*\sqrt{7}=2\sqrt{7}$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{7}}{2*6}=\frac{-2-2\sqrt{7}}{12} $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{7}}{2*6}=\frac{-2+2\sqrt{7}}{12} $
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