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3c^2+19c+28=0
a = 3; b = 19; c = +28;
Δ = b2-4ac
Δ = 192-4·3·28
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{25}=5$$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-5}{2*3}=\frac{-24}{6} =-4 $$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+5}{2*3}=\frac{-14}{6} =-2+1/3 $
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