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15r^2+19r-10=0
a = 15; b = 19; c = -10;
Δ = b2-4ac
Δ = 192-4·15·(-10)
Δ = 961
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{961}=31$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-31}{2*15}=\frac{-50}{30} =-1+2/3 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+31}{2*15}=\frac{12}{30} =2/5 $
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