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25a^2-35a+6=0
a = 25; b = -35; c = +6;
Δ = b2-4ac
Δ = -352-4·25·6
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$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-35)-25}{2*25}=\frac{10}{50} =1/5 $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-35)+25}{2*25}=\frac{60}{50} =1+1/5 $
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