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3r^2-19r+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:$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{25}=5$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-5}{2*3}=\frac{14}{6} =2+1/3 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+5}{2*3}=\frac{24}{6} =4 $
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