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8x^2+32x-96=0
a = 8; b = 32; c = -96;
Δ = b2-4ac
Δ = 322-4·8·(-96)
Δ = 4096
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{4096}=64$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-64}{2*8}=\frac{-96}{16} =-6 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+64}{2*8}=\frac{32}{16} =2 $
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