11x^2-620x+320=0

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Solution for 11x^2-620x+320=0 equation:



11x^2-620x+320=0
a = 11; b = -620; c = +320;
Δ = b2-4ac
Δ = -6202-4·11·320
Δ = 370320
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:
$\sqrt{\Delta}=\sqrt{370320}=\sqrt{16*23145}=\sqrt{16}*\sqrt{23145}=4\sqrt{23145}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-620)-4\sqrt{23145}}{2*11}=\frac{620-4\sqrt{23145}}{22} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-620)+4\sqrt{23145}}{2*11}=\frac{620+4\sqrt{23145}}{22} $

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