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23x^2-75x+50=0
a = 23; b = -75; c = +50;
Δ = b2-4ac
Δ = -752-4·23·50
Δ = 1025
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1025}=\sqrt{25*41}=\sqrt{25}*\sqrt{41}=5\sqrt{41}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-75)-5\sqrt{41}}{2*23}=\frac{75-5\sqrt{41}}{46} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-75)+5\sqrt{41}}{2*23}=\frac{75+5\sqrt{41}}{46} $
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