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25=42x-x^2
We move all terms to the left:
25-(42x-x^2)=0
We get rid of parentheses
x^2-42x+25=0
a = 1; b = -42; c = +25;
Δ = b2-4ac
Δ = -422-4·1·25
Δ = 1664
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1664}=\sqrt{64*26}=\sqrt{64}*\sqrt{26}=8\sqrt{26}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-42)-8\sqrt{26}}{2*1}=\frac{42-8\sqrt{26}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-42)+8\sqrt{26}}{2*1}=\frac{42+8\sqrt{26}}{2} $
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