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32x^2+480x-576=0
a = 32; b = 480; c = -576;
Δ = b2-4ac
Δ = 4802-4·32·(-576)
Δ = 304128
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{304128}=\sqrt{9216*33}=\sqrt{9216}*\sqrt{33}=96\sqrt{33}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(480)-96\sqrt{33}}{2*32}=\frac{-480-96\sqrt{33}}{64} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(480)+96\sqrt{33}}{2*32}=\frac{-480+96\sqrt{33}}{64} $
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