How do you solve #5x ^ { 2} + 3x - 55= 5#?

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Answer 1 · 2017-03-07T05:21:47

#(-3+sqrt1209)/10# or #(-3-sqrt1209)/10#

#5x^2 + 3x - 55 = 5#

=#5x^2 + 3x - 60 = 0#

standard equation of quadratic equation,
#ax^2 + bx + c = 0#

Here,
a = 5, b = 3 and c = 60

by formula,
x = #[-b +- sqrt(b^2 - 4ac)]/(2a)#

So, x = #[-3 +- sqrt(9 + 1200)]/(10)#

#therefore# x = #(-3+sqrt1209)/10# or #(-3-sqrt1209)/10#