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0.5a^2+4=a^2
We move all terms to the left:
0.5a^2+4-(a^2)=0
We add all the numbers together, and all the variables
-0.5a^2+4=0
a = -0.5; b = 0; c = +4;
Δ = b2-4ac
Δ = 02-4·(-0.5)·4
Δ = 8
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}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{8}=\sqrt{4*2}=\sqrt{4}*\sqrt{2}=2\sqrt{2}$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{2}}{2*-0.5}=\frac{0-2\sqrt{2}}{-1} =-\frac{2\sqrt{2}}{-1} $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{2}}{2*-0.5}=\frac{0+2\sqrt{2}}{-1} =\frac{2\sqrt{2}}{-1} $
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