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0.5x^2-46=0
a = 0.5; b = 0; c = -46;
Δ = b2-4ac
Δ = 02-4·0.5·(-46)
Δ = 92
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{92}=\sqrt{4*23}=\sqrt{4}*\sqrt{23}=2\sqrt{23}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{23}}{2*0.5}=\frac{0-2\sqrt{23}}{1} =-\frac{2\sqrt{23}}{1} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{23}}{2*0.5}=\frac{0+2\sqrt{23}}{1} =\frac{2\sqrt{23}}{1} $
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