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