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44x^2+40x-4=0
a = 44; b = 40; c = -4;
Δ = b2-4ac
Δ = 402-4·44·(-4)
Δ = 2304
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{2304}=48$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-48}{2*44}=\frac{-88}{88} =-1 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+48}{2*44}=\frac{8}{88} =1/11 $
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