How do you solve #4x^2 +4x=--11#?

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Answer 1 · 2018-05-29T16:23:11

#x = (-4 + sqrt192)/8 or x = (-4 - sqrt192)/8#

#4x^2 + 4x = - - 11#

#4x^2 + 4x = - (-11)#, Note: #- xx - = +#

#4x^2 + 4x = 11#

#4x^2 + 4x - 11 = 0#

Using;

#x = (-b +- sqrt(b^2 - 4ac))/(2a)#

Where;

#a = 4#

#b = 4#

#c = -11#

Substituting the values of , #a, b, c#

#x = (-4 +- sqrt(4^2 - 4(4)(-11)))/(2(4)#

#x = (-4 +- sqrt(16 +176))/8#

#x = (-4 +- sqrt192)/8#

#x = (-4 + sqrt192)/8 or x = (-4 - sqrt192)/8#