How do you solve the quadratic using the quadratic formula given #b^2-4b-14=-2#?

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Answer 1 · 2016-08-12T16:22:23

Let the equation be #x^2 - 4x - 14 = -2# so that the variables in our equation don't conflict with the parameters in the quadratic formula.

The quadratic formula is #x = (-b +- sqrt(b^2 - 4ac))/(2a)#, for equations of the form #ax^2 + bx + c = 0#.

Sending all our terms to one side:

#x^2 - 4x - 14 = -2#

#x^2 - 4x - 14 + 2 = 0#

#x^2 - 4x - 12 = 0#

Applying the formula:

#x = (-(-4) +- sqrt((-4)^2 - (4 xx 1 xx -12)))/(2 xx 1)#

#x = (4 +- sqrt(16+ 48))/2#

#x = (4 +- sqrt(64))/2#

#x = (4 +- 8) /2#

#x = (4 + 8)/2 and (4 - 8)/2#

#x = 12/2 and -4/2#

#x = 6 and -2#

Checking in the original equation , we have that both solutions work!

Therefore, the solution set is #{x = 6" & "-2}#.

Hopefully this helps!