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(2/5)x=31
We move all terms to the left:
(2/5)x-(31)=0
Domain of the equation: 5)x!=0We add all the numbers together, and all the variables
x!=0/1
x!=0
x∈R
(+2/5)x-31=0
We multiply parentheses
2x^2-31=0
a = 2; b = 0; c = -31;
Δ = b2-4ac
Δ = 02-4·2·(-31)
Δ = 248
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{248}=\sqrt{4*62}=\sqrt{4}*\sqrt{62}=2\sqrt{62}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{62}}{2*2}=\frac{0-2\sqrt{62}}{4} =-\frac{2\sqrt{62}}{4} =-\frac{\sqrt{62}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{62}}{2*2}=\frac{0+2\sqrt{62}}{4} =\frac{2\sqrt{62}}{4} =\frac{\sqrt{62}}{2} $
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