Answer:
The solutions are:
x = -2.11 or 1.11
Explanation:
Given the equation:
[tex]3x^2=-3x+7[/tex]
This can be written as:
[tex]3x^2+3x-7=0[/tex]
Comparing this with the general equation;
[tex]ax^2+bx+c=0[/tex]
We see that;
[tex]\begin{gathered} a=3 \\ b=3 \\ c=-7 \end{gathered}[/tex]
The quadratic formula is:
[tex]x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]
Substitute the values of a, b, and c
[tex]\begin{gathered} x=\frac{-3\pm\sqrt[]{3^2-4\times3\times(-7)}_{}}{2\times3} \\ \\ =\frac{-3\pm\sqrt[]{9+84}}{6} \\ \\ =\frac{-3\pm\sqrt[]{93}}{6} \\ \\ =\frac{-3\pm9.64}{6} \\ \\ x=\frac{-3+9.64}{6}=1.11 \\ \\ OR \\ x=\frac{-3-9.64}{6}=-2.11 \end{gathered}[/tex]
