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8x^2-40x-48=0
a = 8; b = -40; c = -48;
Δ = b2-4ac
Δ = -402-4·8·(-48)
Δ = 3136
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{3136}=56$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-56}{2*8}=\frac{-16}{16} =-1 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+56}{2*8}=\frac{96}{16} =6 $
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