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x^2+2x+3=66
We move all terms to the left:
x^2+2x+3-(66)=0
We add all the numbers together, and all the variables
x^2+2x-63=0
a = 1; b = 2; c = -63;
Δ = b2-4ac
Δ = 22-4·1·(-63)
Δ = 256
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{256}=16$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-16}{2*1}=\frac{-18}{2} =-9 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+16}{2*1}=\frac{14}{2} =7 $
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