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16x^2+64x-132=0
a = 16; b = 64; c = -132;
Δ = b2-4ac
Δ = 642-4·16·(-132)
Δ = 12544
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{12544}=112$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(64)-112}{2*16}=\frac{-176}{32} =-5+1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(64)+112}{2*16}=\frac{48}{32} =1+1/2 $
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