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8b^2+32b-40=0
a = 8; b = 32; c = -40;
Δ = b2-4ac
Δ = 322-4·8·(-40)
Δ = 2304
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{2304}=48$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-48}{2*8}=\frac{-80}{16} =-5 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+48}{2*8}=\frac{16}{16} =1 $
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