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16a^2-25a=0
a = 16; b = -25; c = 0;
Δ = b2-4ac
Δ = -252-4·16·0
Δ = 625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{625}=25$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-25)-25}{2*16}=\frac{0}{32} =0 $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+25}{2*16}=\frac{50}{32} =1+9/16 $
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