What I would do is

let y=x^2

so y^2 – 7y – 3=0

but that doesn’t factorise. so you use the weird formula to determine the value of y

in ay^2+by+c

y= +or-b

in ay^2+by+c

y= b +or- ROOT(b^2-4ac) all over 2a

in this instance,

y^2 – 7y – 3=0

y= -7 +or- ROOT( (-7)^2- 4*1*-3 all over 2*1

= -7 +or- ROOT (49-12) all over 2

= -7 +or- ROOT37 all over 2

and then y=x^2 so it’s the root of all that.

Bleugh. that looks awful. i think i got something wrong there…

First 2 answerers incorrect! enable u = x^2. we now have u^2 – 7u – 3 = 0 that’s no longer factorable, so we use the quadratic formulation: u = (-b +- sqrt(b^2 – 4ac))/2a, the place a = a million, b = -7, c = -3 u = (7 +- sqrt(40 9 – 4(a million)(-3)) )/2(a million) = (7 +- sqrt(sixty one)/2, and the two solutions are u = (7 + sqrt(sixty one))/2, that’s helpful and u = (7-sqrt(sixty one))/2, that’s destructive. because of the fact u = x^2, the only solutions are x = + – sqrt( (7 + sqrt(sixty one))/2), verify the two the helpful and destructive case, and the two ought to paintings. sturdy luck.