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23x^2+2x-44=0
a = 23; b = 2; c = -44;
Δ = b2-4ac
Δ = 22-4·23·(-44)
Δ = 4052
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{4052}=\sqrt{4*1013}=\sqrt{4}*\sqrt{1013}=2\sqrt{1013}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{1013}}{2*23}=\frac{-2-2\sqrt{1013}}{46} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{1013}}{2*23}=\frac{-2+2\sqrt{1013}}{46} $
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