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24a^2+13a-2=0
a = 24; b = 13; c = -2;
Δ = b2-4ac
Δ = 132-4·24·(-2)
Δ = 361
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{361}=19$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-19}{2*24}=\frac{-32}{48} =-2/3 $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+19}{2*24}=\frac{6}{48} =1/8 $
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