Answer:
\(x=1/6\) or \(x=-4\)
Step-by-step explanation:
we know that
The formula to solve a quadratic equation of the form \(ax^{2} +bx+c=0\) is equal to
\(x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}\)
in this problem we have
\(g(x)=6x^{2} +23x-4\)
equate the function to zero
\(6x^{2} +23x-4=0\)
so
\(a=6\\b=23\\c=-4\)
substitute in the formula
\(x=\frac{-23(+/-)\sqrt{23^{2}-4(6)(-4)}} {2(6)}\)
\(x=\frac{-23(+/-)\sqrt{529+96}} {12}\)
\(x=\frac{-23(+/-)25} {12}\)
\(x1=\frac{-23+25} {12}=1/6\)
\(x2=\frac{-23-25} {12}=-4\)
The solution is \(x=1/6\) or \(x=-4\)
