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3x^2+1800x-560=0
a = 3; b = 1800; c = -560;
Δ = b2-4ac
Δ = 18002-4·3·(-560)
Δ = 3246720
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{3246720}=\sqrt{64*50730}=\sqrt{64}*\sqrt{50730}=8\sqrt{50730}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1800)-8\sqrt{50730}}{2*3}=\frac{-1800-8\sqrt{50730}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1800)+8\sqrt{50730}}{2*3}=\frac{-1800+8\sqrt{50730}}{6} $
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