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15r^2-4r-3=0
a = 15; b = -4; c = -3;
Δ = b2-4ac
Δ = -42-4·15·(-3)
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$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-14}{2*15}=\frac{-10}{30} =-1/3 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+14}{2*15}=\frac{18}{30} =3/5 $
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