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3r^2-4r-15=0
a = 3; b = -4; c = -15;
Δ = b2-4ac
Δ = -42-4·3·(-15)
Δ = 196
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{196}=14$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-14}{2*3}=\frac{-10}{6} =-1+2/3 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+14}{2*3}=\frac{18}{6} =3 $
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