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(1/5)x-40=2x-4
We move all terms to the left:
(1/5)x-40-(2x-4)=0
Domain of the equation: 5)x!=0We add all the numbers together, and all the variables
x!=0/1
x!=0
x∈R
(+1/5)x-(2x-4)-40=0
We multiply parentheses
x^2-(2x-4)-40=0
We get rid of parentheses
x^2-2x+4-40=0
We add all the numbers together, and all the variables
x^2-2x-36=0
a = 1; b = -2; c = -36;
Δ = b2-4ac
Δ = -22-4·1·(-36)
Δ = 148
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{148}=\sqrt{4*37}=\sqrt{4}*\sqrt{37}=2\sqrt{37}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{37}}{2*1}=\frac{2-2\sqrt{37}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{37}}{2*1}=\frac{2+2\sqrt{37}}{2} $
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