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16x^2+32x-128=0
a = 16; b = 32; c = -128;
Δ = b2-4ac
Δ = 322-4·16·(-128)
Δ = 9216
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{9216}=96$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-96}{2*16}=\frac{-128}{32} =-4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+96}{2*16}=\frac{64}{32} =2 $
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