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25a^2-25a+6=0
a = 25; b = -25; c = +6;
Δ = b2-4ac
Δ = -252-4·25·6
Δ = 25
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{25}=5$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-25)-5}{2*25}=\frac{20}{50} =2/5 $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+5}{2*25}=\frac{30}{50} =3/5 $
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