uh, where's teh 'given' quadratic equation?
Assuming that the quadratic equation is ax^2 + bx + c = 0, this is the method:
a(x^2 + bx/a) + c = 0
a(x^2 + bx/a + (b^2)/4a) = (b^2)/4a - c
a(x + b/2a)^2 = (b^2 - 4ac)/4a
(x + b/2a)^2 = (b^2 - 4ac)/4a^2
taking the ssquareroot fo both sides:
x + b/2a = +=(√(b²-4ac))/2a
x = -b +- (√(b²-4ac))/2a
just ends up like the quadratic equation, really... well, the quadratic equation was derived somewhat similarly, anyways.. so just substitute the numbers on the variables and you'd be good to go
