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16x^2+20x-104=0
a = 16; b = 20; c = -104;
Δ = b2-4ac
Δ = 202-4·16·(-104)
Δ = 7056
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{7056}=84$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-84}{2*16}=\frac{-104}{32} =-3+1/4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+84}{2*16}=\frac{64}{32} =2 $
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