SOLUTION: hello, my problem here is as follows; 10+8q=3q^2 =(-3q)^2+8q+10 = -(-3q)+/- squared root of (-3)^2-4X(-3)X10/2X(-3) {9-(-120)} =-(-3) +/- squared root of 129/-6 = -21.5

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons -> SOLUTION: hello, my problem here is as follows; 10+8q=3q^2 =(-3q)^2+8q+10 = -(-3q)+/- squared root of (-3)^2-4X(-3)X10/2X(-3) {9-(-120)} =-(-3) +/- squared root of 129/-6 = -21.5       Log On


   

Question 430443: hello, my problem here is as follows;
10+8q=3q^2
=(-3q)^2+8q+10
= -(-3q)+/- squared root of (-3)^2-4X(-3)X10/2X(-3)
{9-(-120)}
=-(-3) +/- squared root of 129/-6
= -21.5
i am stuck at one of these steps that i have done (i am unsure of which one)
because a negative integer cannot be square rooted.
please help me, thank you very much for your time and effort, it is greatly appreciated!
sincerely,
lost 15 year old
Answer by mananth(16949) About Me  (Show Source):
You can put this solution on YOUR website!
10+8q=3q^2
rewrite the equation in ax^2 +bx +c form
3q^2-8q-10
compare with ax^2+bx+c
a= 3 ,b= -8 ,c= -10

b^2-4ac= 64 + 120
b^2-4ac= 184
q=%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F%282a%29
q1=%28-b%2Bsqrt%28b%5E2-4ac%29%29%2F%282a%29
q1=%288%2Bsqrt%28184%29%29%2F%286%29
q1=(8+13.56)/6
q1=3.59
q2=(8-13.56)/6
q2= -0.93