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(x^2)-25x+32=0
a = 1; b = -25; c = +32;
Δ = b2-4ac
Δ = -252-4·1·32
Δ = 497
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-25)-\sqrt{497}}{2*1}=\frac{25-\sqrt{497}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+\sqrt{497}}{2*1}=\frac{25+\sqrt{497}}{2} $
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