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2x-3x^2=-1
We move all terms to the left:
2x-3x^2-(-1)=0
We add all the numbers together, and all the variables
-3x^2+2x+1=0
a = -3; b = 2; c = +1;
Δ = b2-4ac
Δ = 22-4·(-3)·1
Δ = 16
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{16}=4$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-4}{2*-3}=\frac{-6}{-6} =1 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+4}{2*-3}=\frac{2}{-6} =-1/3 $
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