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3r^2+14r-49=0
a = 3; b = 14; c = -49;
Δ = b2-4ac
Δ = 142-4·3·(-49)
Δ = 784
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{784}=28$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-28}{2*3}=\frac{-42}{6} =-7 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+28}{2*3}=\frac{14}{6} =2+1/3 $
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