Question 625471
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Step 1:  Add the opposite of the constant term to both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4x^2\ +\ 32x\ =\ -20]


Step 2:  Multiply both sides by the reciprocal of the lead coefficient:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1}{4}\left(4x^2\ +\ 32x\right)\ =\ \frac{1}{4}\left(-20\right)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ 8x\ =\ -5]


Step 3:  Divide the first degree term coefficient by 2 and square the result.  Add that result to both sides: (8 divided by 2 is 4, 4 squared is 16)


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ 8x\ +\ 16\ =\ 11]


Step 4:  Factor the perfect square trinomial in the RHS.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\ +\ 4)^2\ =\ 11]


Step 5:  Take the square root of both sides, remembering to consider both the positive and negative roots.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ +\ 4\ =\ \pm\sqrt{11}]


Step 6:  Add the opposite of the constant term in the LHS to both sides.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ -4\ \pm\ sqrt{11}]


Step 7:  Send your lifesaving tutor a large gratuity.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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