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2/5a^2+3=11
We move all terms to the left:
2/5a^2+3-(11)=0
Domain of the equation: 5a^2!=0We add all the numbers together, and all the variables
a^2!=0/5
a^2!=√0
a!=0
a∈R
2/5a^2-8=0
We multiply all the terms by the denominator
-8*5a^2+2=0
Wy multiply elements
-40a^2+2=0
a = -40; b = 0; c = +2;
Δ = b2-4ac
Δ = 02-4·(-40)·2
Δ = 320
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{320}=\sqrt{64*5}=\sqrt{64}*\sqrt{5}=8\sqrt{5}$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{5}}{2*-40}=\frac{0-8\sqrt{5}}{-80} =-\frac{8\sqrt{5}}{-80} =-\frac{\sqrt{5}}{-10} $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{5}}{2*-40}=\frac{0+8\sqrt{5}}{-80} =\frac{8\sqrt{5}}{-80} =\frac{\sqrt{5}}{-10} $
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