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8x^2-28x-16=0
a = 8; b = -28; c = -16;
Δ = b2-4ac
Δ = -282-4·8·(-16)
Δ = 1296
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{1296}=36$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-36}{2*8}=\frac{-8}{16} =-1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+36}{2*8}=\frac{64}{16} =4 $
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