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4x^2-8x-96=0
a = 4; b = -8; c = -96;
Δ = b2-4ac
Δ = -82-4·4·(-96)
Δ = 1600
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{1600}=40$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-40}{2*4}=\frac{-32}{8} =-4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+40}{2*4}=\frac{48}{8} =6 $
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