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3x^2-4x-1440=0
a = 3; b = -4; c = -1440;
Δ = b2-4ac
Δ = -42-4·3·(-1440)
Δ = 17296
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{17296}=\sqrt{16*1081}=\sqrt{16}*\sqrt{1081}=4\sqrt{1081}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{1081}}{2*3}=\frac{4-4\sqrt{1081}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{1081}}{2*3}=\frac{4+4\sqrt{1081}}{6} $
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