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