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x=3.x^2+6x
We move all terms to the left:
x-(3.x^2+6x)=0
We get rid of parentheses
-3.x^2+x-6x=0
We add all the numbers together, and all the variables
-3x^2-5x=0
a = -3; b = -5; c = 0;
Δ = b2-4ac
Δ = -52-4·(-3)·0
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-5}{2*-3}=\frac{0}{-6} =0 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+5}{2*-3}=\frac{10}{-6} =-1+2/3 $
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