how do you solve r^2-4r-3=6?

Question

i need help with my math home work

Answer 1

r^2-4r-3=6

r^2 -4r -9 = 0 (subtract 6 from both sides)

Following the quadratic formula of (-b +/- square root of (b^2 -4ac)) / (2a), where b=-4, a=1, and c=-9 in this particular equation:

r= (-(-4) +/- square root of ((-4)^2 - 4(1)(-9)))/(2(1))

r= (4+/- s.r.(16+36))/2        (s.r. means square root)

r₁= 2+  s.r.(13) and r₂= 2-  s.r.(13)  

There are two answers for this problem.  Unfortunately, this is as clear as I can type this.  If you have any questions, please let me know.  This answer box doesn't allow me to use actual math typing.  Green font indicates the answers, orange font indicates the steps.

Answer 2

r² - 4r - 3 = 6

Normally, we can solve by factoring the equation (if factorable) and setting each factor equal to zero. First we need to get the equation in standard form (Ax² + Bx + C = 0) by subtracting 6 from each side:

r² - 4r - 9 = 0

Now, we factor the trinomial into the product of two binomials - however, this is not factorable so we use the quadratic equation:

 r =   -b +/- √(b² - 4ac)
                  2a

where a = 1, b = -4 are coefficients of the quadratic and linear terms, respectively, and c = -9 is the constant term (see standard form of equation above).

Now just plug the values into the formula and evaluate. You should get two answers:

r = 2 + sqrt(13), and 2 - sqrt(13)