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3x^2=2+x
We move all terms to the left:
3x^2-(2+x)=0
We add all the numbers together, and all the variables
3x^2-(x+2)=0
We get rid of parentheses
3x^2-x-2=0
We add all the numbers together, and all the variables
3x^2-1x-2=0
a = 3; b = -1; c = -2;
Δ = b2-4ac
Δ = -12-4·3·(-2)
Δ = 25
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{25}=5$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-5}{2*3}=\frac{-4}{6} =-2/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+5}{2*3}=\frac{6}{6} =1 $
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