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0.5x^2-9=0
a = 0.5; b = 0; c = -9;
Δ = b2-4ac
Δ = 02-4·0.5·(-9)
Δ = 18
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{18}=\sqrt{9*2}=\sqrt{9}*\sqrt{2}=3\sqrt{2}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-3\sqrt{2}}{2*0.5}=\frac{0-3\sqrt{2}}{1} =-\frac{3\sqrt{2}}{1} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+3\sqrt{2}}{2*0.5}=\frac{0+3\sqrt{2}}{1} =\frac{3\sqrt{2}}{1} $
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