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2x^2-2+3x+3=0
We add all the numbers together, and all the variables
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-1}{2*2}=\frac{-4}{4} =-1 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+1}{2*2}=\frac{-2}{4} =-1/2 $
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