Given:
The equation is given as x²+4x = -13.
The objective is to solve the quadratic formula.
Explanation:
The equation can be rewritten as,
[tex]x^2+4x+13=0\text{ . . . . . . (1)}[/tex]
Consider the coefficients of the equation as,
[tex]\begin{gathered} a=1 \\ b=4 \\ c=13 \end{gathered}[/tex]
To find solution:
The quadratic formula to find the solutions is,
[tex]x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}\text{ . . . . . .(1)}[/tex]
On plugging the obtained values in equation (1),
[tex]\begin{gathered} x=\frac{-4\pm\sqrt[]{(-4)^2-4(1)(13)}}{2(1)} \\ x=\frac{-4\pm\sqrt[]{16-52}}{2} \\ x=\frac{-4\pm\sqrt[]{-36}}{2} \end{gathered}[/tex]
Since, the discriminant is less than 1, the solutions will be complex roots.
[tex]\sqrt[\square]{(-1)}=i[/tex]
On further solving the above equation,
[tex]\begin{gathered} x=\frac{-4\pm6i}{2} \\ x=\frac{-4}{2}\pm\frac{6i}{2} \\ x=-2\pm3i \\ x=-2+3i,-2-3i \end{gathered}[/tex]
Hence, the solutions of the equation are (-2+3i) and (-2-3i).
