How do you find the solution to the quadratic equation #x^2 + 8x - 3 = 0#?

1 Answer
George C.
May 14, 2015

#x^2+8x-3 = 0# is of the form #ax^2+bx+c = 0# with #a=1#, #b=8# and #c=-3#.

The discriminant:

#Delta = b^2-4ac = 8^2-4*1*(-3) = 64 + 12 = 76#.

Since #Delta > 0#, the quadratic equation has 2 distinct real solutions. It is not a perfect square, so those solutions will not be integers or rational numbers.

The general formula to solve a quadratic is:

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

In our case:

#x = (-8 +- sqrt(76))/2 = (-8 += sqrt(4*19))/2 = (-8 +- 2sqrt(19))/2#

#= -4 +- sqrt(19)#

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