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8b^2+4b-40=0
a = 8; b = 4; c = -40;
Δ = b2-4ac
Δ = 42-4·8·(-40)
Δ = 1296
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{1296}=36$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-36}{2*8}=\frac{-40}{16} =-2+1/2 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+36}{2*8}=\frac{32}{16} =2 $
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