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4x^2-560x+3=0
a = 4; b = -560; c = +3;
Δ = b2-4ac
Δ = -5602-4·4·3
Δ = 313552
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{313552}=\sqrt{16*19597}=\sqrt{16}*\sqrt{19597}=4\sqrt{19597}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-560)-4\sqrt{19597}}{2*4}=\frac{560-4\sqrt{19597}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-560)+4\sqrt{19597}}{2*4}=\frac{560+4\sqrt{19597}}{8} $
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