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2x^2+2x=3x+6
We move all terms to the left:
2x^2+2x-(3x+6)=0
We get rid of parentheses
2x^2+2x-3x-6=0
We add all the numbers together, and all the variables
2x^2-1x-6=0
a = 2; b = -1; c = -6;
Δ = b2-4ac
Δ = -12-4·2·(-6)
Δ = 49
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{49}=7$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-7}{2*2}=\frac{-6}{4} =-1+1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+7}{2*2}=\frac{8}{4} =2 $
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