Question

Enter the correct answer in the box.What are the solutions of this quadratic equation?22 - 61 = -58Substitute the values of a and b to complete the solutions.OP+TT0 0 0vo yoα β ε 9sin cos tan sin cos'tan-1A1 -Io =tanncoXV?-WEB<ρ φΣ00L002CSC seccot lag log in1Ux= a + bix=a-bi.ResetNextmentum. All rights reserved

Answer 1

The solutions to a quadratic equation are found using the quadratic formula. In one version provided, the equation is x² + 0.0211x - 0.0211 = 0 with a = 1, b = 0.0211, c = -0.0211. Substituting these values into the formula yields the solutions.

Step-by-step explanation:

The solutions to a quadratic equation of the form ax² + bx + c = 0 can be found using the quadratic formula:

x = −b ± √(b² − 4ac) / (2a)

It seems that the question contains multiple typos and variations of the quadratic equation, but one version provided includes:

a = 1, b = 0.0211, and c = -0.0211. Substituting these values into the quadratic formula gives:

x = −(0.0211) ± √((0.0211)² − 4(1)(−0.0211)) / (2(1))

A calculation using these values will yield the solutions to the equation.

Answer 2

The solution to the given equation is;

`\begin{gathered} x=3+7i \\ x=3-7i \end{gathered}`

Step-by-step explanation:

Given the quadratic equation;

`x^2-6x=-58`

Solving the quadratic equation by applying the quadratic formula;

`x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}`

Given the quadratic equation as;

`\begin{gathered} x^2-6x=-58 \\ x^2-6x+58=0 \\ a=1 \\ b=-6 \\ c=58 \end{gathered}`

substituting;

`\begin{gathered} x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ x=\frac{6\pm\sqrt[]{(-6)^2-4(1)(58)}}{2(1)} \\ x=\frac{6\pm\sqrt[]{36^{}-4(1)(58)}}{2(1)} \end{gathered}`
`\begin{gathered} x=\frac{6\pm\sqrt[]{36^{}-232}}{2} \\ x=\frac{6\pm\sqrt[]{-196}}{2} \\ x=(6\pm14i)/(2) \\ x=3\pm7i \end{gathered}`

Therefore, the solution to the given equation is;

`\begin{gathered} x=3+7i \\ x=3-7i \end{gathered}`