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24x^2-4x-3=0
a = 24; b = -4; c = -3;
Δ = b2-4ac
Δ = -42-4·24·(-3)
Δ = 304
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{304}=\sqrt{16*19}=\sqrt{16}*\sqrt{19}=4\sqrt{19}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{19}}{2*24}=\frac{4-4\sqrt{19}}{48} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{19}}{2*24}=\frac{4+4\sqrt{19}}{48} $
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