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49r^2-16r=0
a = 49; b = -16; c = 0;
Δ = b2-4ac
Δ = -162-4·49·0
Δ = 256
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{256}=16$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-16}{2*49}=\frac{0}{98} =0 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+16}{2*49}=\frac{32}{98} =16/49 $
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