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2x^2+2x+12=3x^2-x+2
We move all terms to the left:
2x^2+2x+12-(3x^2-x+2)=0
We get rid of parentheses
2x^2-3x^2+2x+x-2+12=0
We add all the numbers together, and all the variables
-1x^2+3x+10=0
a = -1; b = 3; c = +10;
Δ = b2-4ac
Δ = 32-4·(-1)·10
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-7}{2*-1}=\frac{-10}{-2} =+5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+7}{2*-1}=\frac{4}{-2} =-2 $
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