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-16x^2-160x+32=0
a = -16; b = -160; c = +32;
Δ = b2-4ac
Δ = -1602-4·(-16)·32
Δ = 27648
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{27648}=\sqrt{9216*3}=\sqrt{9216}*\sqrt{3}=96\sqrt{3}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-160)-96\sqrt{3}}{2*-16}=\frac{160-96\sqrt{3}}{-32} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-160)+96\sqrt{3}}{2*-16}=\frac{160+96\sqrt{3}}{-32} $
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