x^(2)-13 x+30=0 What are the solutions to the given equation? Choose 1 answer: (A) x=-15 and x=2 (B) x=-10 and x=-3 (C) x=3 and x=10 (D) x=-2 and x=15

Given equation: The given equation is a quadratic equation of the form ax^2 + bx + c = 0. We need to find the values of x that satisfy the equation x^2 - 13x + 30 = 0. To do this, we can factor the quadratic if possible.Factoring the quadratic: We look for two numbers that multiply to give the constant term, c = 30, and add up to give the coefficient of the x term, b = -13. The numbers that satisfy these conditions are -3 and -10 because (-3) * (-10) = 30 and (-3) + (-10) = -13.Setting up the equations: We can now factor the quadratic equation as (x - 3)(x - 10) = 0.Solving the first equation: To find the solutions, we set each factor equal to zero and solve for x. This gives us two equations: x - 3 = 0 and x - 10 = 0.Solving the second equation: Solving the first equation, x - 3 = 0, we get x = 3.Solution to the quadratic equation: Solving the second equation, x - 10 = 0, we get x = 10.We have found the two solutions to the quadratic equation: x = 3 and x = 10. These are the values of x that satisfy the equation x^2 - 13x + 30 = 0.

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x^(2)-13 x+30=0
What are the solutions to the given equation?
Choose 1 answer:
(A)
x=-15 and
x=2
(B)
x=-10 and
x=-3
(C)
x=3 and
x=10
(D)
x=-2 and
x=15

$x^{2}-13 x+30=0$ $\newline$ What are the solutions to the given equation? $\newline$ Choose $1$ answer: $\newline$ (A) $x=-15$ and $x=2$ $\newline$ (B) $x=-10$ and $x=-3$ $\newline$ (C) $x=3$ and $x=10$ $\newline$ (D) $x=-2$ and $x=15$

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Full solution

x^{2}-13 x+30=0

\newline

What are the solutions to the given equation?

\newline

Choose

1

answer:

\newline

(A)

x=-15

and

x=2

\newline

(B)

x=-10

and

x=-3

\newline

(C)

x=3

and

x=10

\newline

(D)

x=-2

and

x=15

Given equation: The given equation is a quadratic equation of the form $ax^2 + bx + c = 0$ . We need to find the values of $x$ that satisfy the equation $x^2 - 13x + 30 = 0$ . To do this, we can factor the quadratic if possible.
Factoring the quadratic: We look for two numbers that multiply to give the constant term, $c = 30$ , and add up to give the coefficient of the $x$ term, $b = -13$ . The numbers that satisfy these conditions are $-3$ and $-10$ because $(-3) \times (-10) = 30$ and $(-3) + (-10) = -13$ .
Setting up the equations: We can now factor the quadratic equation as $(x - 3)(x - 10) = 0$ .
Solving the first equation: To find the solutions, we set each factor equal to zero and solve for x. This gives us two equations: $x - 3 = 0$ and $x - 10 = 0$ .
Solving the second equation: Solving the first equation, $x - 3 = 0$ , we get $x = 3$ .
Solution to the quadratic equation: Solving the second equation, $x - 10 = 0$ , we get $x = 10$ .
Solution to the quadratic equation: Solving the second equation, $x - 10 = 0$ , we get $x = 10$ . We have found the two solutions to the quadratic equation: $x = 3$ and $x = 10$ . These are the values of $x$ that satisfy the equation $x^2 - 13x + 30 = 0$ .