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6r^2+8r-8=0
a = 6; b = 8; c = -8;
Δ = b2-4ac
Δ = 82-4·6·(-8)
Δ = 256
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{256}=16$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-16}{2*6}=\frac{-24}{12} =-2 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+16}{2*6}=\frac{8}{12} =2/3 $
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