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10b^2-25b+10=0
a = 10; b = -25; c = +10;
Δ = b2-4ac
Δ = -252-4·10·10
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{225}=15$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-25)-15}{2*10}=\frac{10}{20} =1/2 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+15}{2*10}=\frac{40}{20} =2 $
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