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-4b^2+40b-96=0
a = -4; b = 40; c = -96;
Δ = b2-4ac
Δ = 402-4·(-4)·(-96)
Δ = 64
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{64}=8$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-8}{2*-4}=\frac{-48}{-8} =+6 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+8}{2*-4}=\frac{-32}{-8} =+4 $
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