Sure! Let's solve these two equations step by step.
### Equation 1: [tex]\((x + 5)(x - 57) = 0\)[/tex]
This is a factored quadratic equation. To find the solutions, we set each factor equal to zero:
1. [tex]\( x + 5 = 0 \)[/tex]
2. [tex]\( x - 57 = 0 \)[/tex]
Solving these individually:
1. [tex]\( x + 5 = 0 \)[/tex]
[tex]\[
x = -5
\][/tex]
2. [tex]\( x - 57 = 0 \)[/tex]
[tex]\[
x = 57
\][/tex]
So, the solutions to the equation [tex]\((x + 5)(x - 57) = 0\)[/tex] are:
[tex]\[
x = -5 \quad \text{and} \quad x = 57
\][/tex]
### Equation 2: [tex]\(5.14 x^2 - 10 x = 0\)[/tex]
This is a quadratic equation in standard form. We can solve it by factoring or using the quadratic formula. Factoring out the common term [tex]\(x\)[/tex]:
[tex]\[
x(5.14x - 10) = 0
\][/tex]
Setting each factor to zero gives us:
1. [tex]\( x = 0 \)[/tex]
2. [tex]\( 5.14x - 10 = 0 \)[/tex]
[tex]\[
5.14x = 10
\][/tex]
[tex]\[
x = \frac{10}{5.14}
\][/tex]
[tex]\[
x \approx 1.94552529182879
\][/tex]
So, the solutions to the equation [tex]\(5.14 x^2 - 10 x = 0\)[/tex] are:
[tex]\[
x = 0 \quad \text{and} \quad x \approx 1.94552529182879
\][/tex]
### Summary
The solutions are:
1. [tex]\((x + 5)(x - 57) = 0\)[/tex]
[tex]\[
x = -5 \quad \text{and} \quad x = 57
\][/tex]
2. [tex]\(5.14 x^2 - 10 x = 0\)[/tex]
[tex]\[
x = 0 \quad \text{and} \quad x \approx 1.94552529182879
\][/tex]
