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32x^2-58x+11=0
a = 32; b = -58; c = +11;
Δ = b2-4ac
Δ = -582-4·32·11
Δ = 1956
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1956}=\sqrt{4*489}=\sqrt{4}*\sqrt{489}=2\sqrt{489}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-58)-2\sqrt{489}}{2*32}=\frac{58-2\sqrt{489}}{64} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-58)+2\sqrt{489}}{2*32}=\frac{58+2\sqrt{489}}{64} $
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