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