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05x^2+2x+8=8
We move all terms to the left:
05x^2+2x+8-(8)=0
We add all the numbers together, and all the variables
05x^2+2x=0
a = 05; b = 2; c = 0;
Δ = b2-4ac
Δ = 22-4·05·0
Δ = 4
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{4}=2$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2}{2*05}=\frac{-4}{10} =-2/5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2}{2*05}=\frac{0}{10} =0 $
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