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0.5x^2-6x+3=7
We move all terms to the left:
0.5x^2-6x+3-(7)=0
We add all the numbers together, and all the variables
0.5x^2-6x-4=0
a = 0.5; b = -6; c = -4;
Δ = b2-4ac
Δ = -62-4·0.5·(-4)
Δ = 44
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{44}=\sqrt{4*11}=\sqrt{4}*\sqrt{11}=2\sqrt{11}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{11}}{2*0.5}=\frac{6-2\sqrt{11}}{1} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{11}}{2*0.5}=\frac{6+2\sqrt{11}}{1} $
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