To solve the quadratic equation [tex]\(7x^2 + 10x = -3\)[/tex], we start by rearranging the equation so that it takes the standard form [tex]\(ax^2 + bx + c = 0\)[/tex].
First, we add 3 to both sides of the equation to move all terms to one side:
[tex]\[
7x^2 + 10x + 3 = 0
\][/tex]
Now, we have a standard quadratic equation [tex]\(7x^2 + 10x + 3 = 0\)[/tex]. To find the solutions for [tex]\(x\)[/tex], we can use the quadratic formula, which is given by:
[tex]\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
where [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] are the coefficients of the terms [tex]\(x^2\)[/tex], [tex]\(x\)[/tex], and the constant term in the quadratic equation respectively. In this case:
[tex]\[
a = 7, \quad b = 10, \quad c = 3
\][/tex]
Let's calculate the discriminant [tex]\(\Delta\)[/tex] first:
[tex]\[
\Delta = b^2 - 4ac = 10^2 - 4(7)(3) = 100 - 84 = 16
\][/tex]
Now, we substitute [tex]\(a\)[/tex], [tex]\(b\)[/tex], [tex]\(c\)[/tex], and [tex]\(\Delta\)[/tex] into the quadratic formula:
[tex]\[
x = \frac{-b \pm \sqrt{\Delta}}{2a} = \frac{-10 \pm \sqrt{16}}{2(7)} = \frac{-10 \pm 4}{14}
\][/tex]
This will give us two solutions:
[tex]\[
x_1 = \frac{-10 + 4}{14} = \frac{-6}{14} = -\frac{3}{7}
\][/tex]
[tex]\[
x_2 = \frac{-10 - 4}{14} = \frac{-14}{14} = -1
\][/tex]
Therefore, the solutions to the equation [tex]\(7x^2 + 10x + 3 = 0\)[/tex] are:
[tex]\[
x = -1 \quad \text{and} \quad x = -\frac{3}{7}
\][/tex]
