Respuesta :
We are given the following equation
[tex]3x^2+7x+14=-7x[/tex]Let us solve the equation for x.
First of all, rearrange the equation so that all the terms are on the left side of the equation
[tex]\begin{gathered} 3x^2+7x+14=-7x \\ 3x^2+7x+7x+14=0 \\ 3x^2+14x+14=0 \end{gathered}[/tex]Recall that the standard form of a quadratic equation is given by
[tex]ax^2+bx+c=0[/tex]Comparing the given equation with the standard form, the coefficients are
a = 3
b = 14
c = 14
Recall that the quadratic formula is given by
[tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]Let us substitute the values of coefficients and simplify
[tex]\begin{gathered} x=\frac{-14\pm\sqrt[]{14^2-4(3)(14)}}{2(3)} \\ x=\frac{-14\pm\sqrt[]{196^{}-168}}{6} \\ x=\frac{-14\pm\sqrt[]{28}}{6} \\ x=\frac{-14+\sqrt[]{28}}{6},\: \: x=\frac{-14-\sqrt[]{28}}{6} \\ x=-1.5,\: \: x=-3.2 \end{gathered}[/tex]Therefore, the solution of the equation is
[tex]x=\mleft\lbrace-1.5,-3.2\mright\rbrace[/tex]