Question

What is the solution set of the equation using the quadratic formula? x2+6x+10=0 {−3+2i,−3−2i} {−6+2i,−6−2i} {−3+i,−3−i} {−2i,−4i}

Answer 1

Answer:

The solution set of the quadratic function $x^{2}+6\cdot x +10$ is $\{-3+i,-3-i\}$ .

Step-by-step explanation:

Let be a second-order polynomial (quadratic function) is standard form and equalized to zero:

$a\cdot x^{2}+b\cdot x + c = 0$

Its roots can be determined by the Quadratic Formula in terms of its polynomial coefficients, which states that:

$x_{1,2} = \frac{-b\pm\sqrt{b^{2}-4\cdot a\cdot c}}{2\cdot a}$

Given that $a = 1$, $b = 6$ and $c = 10$, the roots of the polynomial are, respectively:

$x_{1,2} = \frac{-6\pm \sqrt{6^{2}-4\cdot (1)\cdot (10)}}{2\cdot (1)}$

$x_{1,2} = -3\pm i$

$x_{1} = -3+i$

$x_{2} = -3 -i$

The solution set of the quadratic function $x^{2}+6\cdot x +10$ is $\{-3+i,-3-i\}$ .