Answer:
Solutions of the quadratic equation are \(\dfrac{2}{3},\dfrac{1}{2}\)
Step-by-step explanation:
The given quadratic equation is,
\(6x^2 - 7x + 2 = 0\)
The quadratic formula is,
\(x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}\)
Here,
\(a = 6,b=-7,c=2\)
Putting the values,
\(x=\dfrac{-(-7)\pm \sqrt{(-7)^2-4\cdot 6\cdot 2}}{2\cdot 6}\)
\(=\dfrac{7\pm \sqrt{49-48}}{12}\)
\(=\dfrac{7\pm \sqrt{1}}{12}\)
\(=\dfrac{7\pm 1}{12}\)
\(=\dfrac{7+ 1}{12},\dfrac{7-1}{12}\)
\(=\dfrac{8}{12},\dfrac{6}{12}\)
\(=\dfrac{2}{3},\dfrac{1}{2}\)
