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0=2x^2+3x+1
We move all terms to the left:
0-(2x^2+3x+1)=0
We add all the numbers together, and all the variables
-(2x^2+3x+1)=0
We get rid of parentheses
-2x^2-3x-1=0
a = -2; b = -3; c = -1;
Δ = b2-4ac
Δ = -32-4·(-2)·(-1)
Δ = 1
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{1}=1$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-1}{2*-2}=\frac{2}{-4} =-1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+1}{2*-2}=\frac{4}{-4} =-1 $
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