Answer:
\(x=-1+\dfrac{\sqrt{33}}{3} , x=-1-\dfrac{\sqrt{33}}{3}\)
Step-by-step explanation:
To solve the equation:
\(3x^{2} +6x=8\)
means we have to find the value of 'x' after solving this quadratic equation
this equation could also be written as:
\(3x^{2}+6x-8=0\)
Also we know that for any quadratic equation of the type:
\(ax^{2}+bx+c=0\)
The solution is given by:
\(x=\dfrac{-b+\sqrt{b^2-4ac}} {2a} ,x=\dfrac{-b-\sqrt{b^2-4ac}} {2a}\)
so on comparing our given equation to the general form we have:
a=3,b=6 and c= -8
so, the value of x is:
\(x=\dfrac{-6+\sqrt{36+32\times3} }{2\times 3} , x=\dfrac{-6-\sqrt{36+32\times3} }{2\times 3}\)
i.e.
\(x=\dfrac{-6+2\sqrt{33} }{6} ,x= \dfrac{-6-2\sqrt{33} }{6}\)
which could also be written as:
\(x=-1+\dfrac{\sqrt{33}}{3} , x=-1-\dfrac{\sqrt{33}}{3}\)
Hence the above two values of x is the desired result.
