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3r^2+32r-48=0
a = 3; b = 32; c = -48;
Δ = b2-4ac
Δ = 322-4·3·(-48)
Δ = 1600
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{1600}=40$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-40}{2*3}=\frac{-72}{6} =-12 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+40}{2*3}=\frac{8}{6} =1+1/3 $
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