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24x^2-98x+99=0
a = 24; b = -98; c = +99;
Δ = b2-4ac
Δ = -982-4·24·99
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{100}=10$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-98)-10}{2*24}=\frac{88}{48} =1+5/6 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-98)+10}{2*24}=\frac{108}{48} =2+1/4 $
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