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3=x^2+3x-9
We move all terms to the left:
3-(x^2+3x-9)=0
We get rid of parentheses
-x^2-3x+9+3=0
We add all the numbers together, and all the variables
-1x^2-3x+12=0
a = -1; b = -3; c = +12;
Δ = b2-4ac
Δ = -32-4·(-1)·12
Δ = 57
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{-(-3)-\sqrt{57}}{2*-1}=\frac{3-\sqrt{57}}{-2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{57}}{2*-1}=\frac{3+\sqrt{57}}{-2} $
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