Certainly! Let's solve the quadratic equation [tex]\(2x^2 + x = 15\)[/tex] step-by-step.
First, we need to bring the equation to a standard quadratic form:
[tex]\[2x^2 + x - 15 = 0\][/tex]
This is a quadratic equation in the form of [tex]\(ax^2 + bx + c = 0\)[/tex], where [tex]\(a = 2\)[/tex], [tex]\(b = 1\)[/tex], and [tex]\(c = -15\)[/tex].
To solve this quadratic equation, we can use the quadratic formula:
[tex]\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\][/tex]
We will plug in the values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] into the quadratic formula:
1. Calculate the discriminant [tex]\( \Delta \)[/tex]:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
[tex]\[ \Delta = 1^2 - 4 \cdot 2 \cdot (-15) \][/tex]
[tex]\[ \Delta = 1 + 120 \][/tex]
[tex]\[ \Delta = 121 \][/tex]
2. Substitute the discriminant [tex]\(\Delta\)[/tex] and the coefficients [tex]\(a\)[/tex], [tex]\(b\)[/tex] into the quadratic formula:
[tex]\[ x = \frac{-1 \pm \sqrt{121}}{2 \cdot 2} \][/tex]
[tex]\[ x = \frac{-1 \pm 11}{4} \][/tex]
3. Solve for the two possible values of [tex]\(x\)[/tex]:
[tex]\[ x_1 = \frac{-1 + 11}{4} \][/tex]
[tex]\[ x_1 = \frac{10}{4} \][/tex]
[tex]\[ x_1 = \frac{5}{2} \][/tex]
[tex]\[ x_2 = \frac{-1 - 11}{4} \][/tex]
[tex]\[ x_2 = \frac{-12}{4} \][/tex]
[tex]\[ x_2 = -3 \][/tex]
Therefore, the solutions to the equation [tex]\(2x^2 + x = 15\)[/tex] are:
[tex]\[ x = \frac{5}{2} \quad \text{and} \quad x = -3 \][/tex]
So, the correct answer is [tex]\(x = \frac{5}{2}\)[/tex] and [tex]\(x = -3\)[/tex].
