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24r^2+15r=0
a = 24; b = 15; c = 0;
Δ = b2-4ac
Δ = 152-4·24·0
Δ = 225
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{225}=15$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-15}{2*24}=\frac{-30}{48} =-5/8 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+15}{2*24}=\frac{0}{48} =0 $
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