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4x^2+2400x-480=0
a = 4; b = 2400; c = -480;
Δ = b2-4ac
Δ = 24002-4·4·(-480)
Δ = 5767680
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{5767680}=\sqrt{256*22530}=\sqrt{256}*\sqrt{22530}=16\sqrt{22530}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2400)-16\sqrt{22530}}{2*4}=\frac{-2400-16\sqrt{22530}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2400)+16\sqrt{22530}}{2*4}=\frac{-2400+16\sqrt{22530}}{8} $
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