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24x^2-35x+12=0
a = 24; b = -35; c = +12;
Δ = b2-4ac
Δ = -352-4·24·12
Δ = 73
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-35)-\sqrt{73}}{2*24}=\frac{35-\sqrt{73}}{48} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-35)+\sqrt{73}}{2*24}=\frac{35+\sqrt{73}}{48} $
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