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3x^2+2x+1=3x+4x^2-3
We move all terms to the left:
3x^2+2x+1-(3x+4x^2-3)=0
We get rid of parentheses
3x^2-4x^2-3x+2x+3+1=0
We add all the numbers together, and all the variables
-1x^2-1x+4=0
a = -1; b = -1; c = +4;
Δ = b2-4ac
Δ = -12-4·(-1)·4
Δ = 17
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-\sqrt{17}}{2*-1}=\frac{1-\sqrt{17}}{-2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{17}}{2*-1}=\frac{1+\sqrt{17}}{-2} $
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