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24x^2+x-10=0
a = 24; b = 1; c = -10;
Δ = b2-4ac
Δ = 12-4·24·(-10)
Δ = 961
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{961}=31$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-31}{2*24}=\frac{-32}{48} =-2/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+31}{2*24}=\frac{30}{48} =5/8 $
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