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-x2+2x+3=-2x^2+4x+6
We move all terms to the left:
-x2+2x+3-(-2x^2+4x+6)=0
We add all the numbers together, and all the variables
-1x^2-(-2x^2+4x+6)+2x+3=0
We get rid of parentheses
-1x^2+2x^2-4x+2x-6+3=0
We add all the numbers together, and all the variables
x^2-2x-3=0
a = 1; b = -2; c = -3;
Δ = b2-4ac
Δ = -22-4·1·(-3)
Δ = 16
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{16}=4$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-4}{2*1}=\frac{-2}{2} =-1 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+4}{2*1}=\frac{6}{2} =3 $
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