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8x^2-48+8x=0
a = 8; b = 8; c = -48;
Δ = b2-4ac
Δ = 82-4·8·(-48)
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-40}{2*8}=\frac{-48}{16} =-3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+40}{2*8}=\frac{32}{16} =2 $
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