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x^2-4x+3=3x+6
We move all terms to the left:
x^2-4x+3-(3x+6)=0
We get rid of parentheses
x^2-4x-3x-6+3=0
We add all the numbers together, and all the variables
x^2-7x-3=0
a = 1; b = -7; c = -3;
Δ = b2-4ac
Δ = -72-4·1·(-3)
Δ = 61
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-\sqrt{61}}{2*1}=\frac{7-\sqrt{61}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+\sqrt{61}}{2*1}=\frac{7+\sqrt{61}}{2} $
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