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3r^2+14r+16=0
a = 3; b = 14; c = +16;
Δ = b2-4ac
Δ = 142-4·3·16
Δ = 4
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{4}=2$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-2}{2*3}=\frac{-16}{6} =-2+2/3 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+2}{2*3}=\frac{-12}{6} =-2 $
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