SOLUTION: Dear Math Tutor, How does 3x^2-5x-6=0 become 1/6(5 plus-minus the square root of 97)and x^2-4x-7=0 become 2 plus-minus the square root of 11? with gratitude, Nat

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Dear Math Tutor, How does 3x^2-5x-6=0 become 1/6(5 plus-minus the square root of 97)and x^2-4x-7=0 become 2 plus-minus the square root of 11? with gratitude, Nat       Log On


   

Question 878375: Dear Math Tutor,
How does 3x^2-5x-6=0 become 1/6(5 plus-minus the square root of 97)and
x^2-4x-7=0 become 2 plus-minus the square root of 11?

with gratitude,
Nat

Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
You can complete the square or use the quadratic formula.
Quadratic formula:
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
x+=+%28-%28-5%29+%2B-+sqrt%28+%28-5%29%5E2-4%2A3%2A%28-6%29%29%29%29%2F%282%2A3%29+
x+=+%285+%2B-+sqrt%28+25%2B72%29%29%2F6+
x+=+%285+%2B-+sqrt%28+97%29%29%2F6+
.
.
.
Completing the square:
x%5E2-4x-7=0
%28x%5E2-4x%2B4%29-7=4
%28x-2%29%5E2-7=4
%28x-2%29%5E2=11
x-2=0+%2B-+sqrt%2811%29
x=2+%2B-+sqrt%2811%29