Question

What are the zeros of the quadratic function f(x)=6x^2+12x-7

Answer 1

Answer:

$A) \ x=-1 \ ^ +_-\frac{\sqrt{13} }{\sqrt{6}}.$

Step-by-step explanation:

if to solve the given equation, then

D=6²+42=78;

$\left[\begin{array}{ccc}x=\frac{-6-\sqrt{78}}{6} \\x=\frac{-6+\sqrt{78}}{6} \end{array} \ = > \ \left[\begin{array}{ccc}x=-1-\sqrt{\frac{13}{6} } \\x=-1+\sqrt{\frac{13}{6} } \end{array}$