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5x^2+128x-240=0
a = 5; b = 128; c = -240;
Δ = b2-4ac
Δ = 1282-4·5·(-240)
Δ = 21184
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{21184}=\sqrt{64*331}=\sqrt{64}*\sqrt{331}=8\sqrt{331}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(128)-8\sqrt{331}}{2*5}=\frac{-128-8\sqrt{331}}{10} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(128)+8\sqrt{331}}{2*5}=\frac{-128+8\sqrt{331}}{10} $
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