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5(x-3)2x=41
We move all terms to the left:
5(x-3)2x-(41)=0
We multiply parentheses
10x^2-30x-41=0
a = 10; b = -30; c = -41;
Δ = b2-4ac
Δ = -302-4·10·(-41)
Δ = 2540
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2540}=\sqrt{4*635}=\sqrt{4}*\sqrt{635}=2\sqrt{635}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-30)-2\sqrt{635}}{2*10}=\frac{30-2\sqrt{635}}{20} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-30)+2\sqrt{635}}{2*10}=\frac{30+2\sqrt{635}}{20} $
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