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12r^2+4r-1=0
a = 12; b = 4; c = -1;
Δ = b2-4ac
Δ = 42-4·12·(-1)
Δ = 64
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}$$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{64}=8$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-8}{2*12}=\frac{-12}{24} =-1/2 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+8}{2*12}=\frac{4}{24} =1/6 $
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