To find the zeros of the quadratic function [tex]\( y = 2x^2 + 7x + 3 \)[/tex], we need to determine the values of [tex]\( x \)[/tex] that make [tex]\( y \)[/tex] equal to zero. That means we are solving the equation:
[tex]\[ 2x^2 + 7x + 3 = 0 \][/tex]
To solve this quadratic equation, we use the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where [tex]\( a = 2 \)[/tex], [tex]\( b = 7 \)[/tex], and [tex]\( c = 3 \)[/tex].
Let's apply the values of [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex] to the quadratic formula:
1. Calculate the discriminant:
[tex]\[ b^2 - 4ac = 7^2 - 4 \cdot 2 \cdot 3 \][/tex]
[tex]\[ = 49 - 24 \][/tex]
[tex]\[ = 25 \][/tex]
2. Substitute the discriminant and the values of [tex]\( a \)[/tex] and [tex]\( b \)[/tex] into the quadratic formula to find the two solutions:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
[tex]\[ x = \frac{-7 \pm \sqrt{25}}{2 \cdot 2} \][/tex]
[tex]\[ x = \frac{-7 \pm 5}{4} \][/tex]
This gives us two possible solutions:
- For the positive root:
[tex]\[ x_1 = \frac{-7 + 5}{4} \][/tex]
[tex]\[ x_1 = \frac{-2}{4} \][/tex]
[tex]\[ x_1 = -0.5 \][/tex]
- For the negative root:
[tex]\[ x_2 = \frac{-7 - 5}{4} \][/tex]
[tex]\[ x_2 = \frac{-12}{4} \][/tex]
[tex]\[ x_2 = -3 \][/tex]
Therefore, the zeros of the function [tex]\( y = 2x^2 + 7x + 3 \)[/tex] are:
[tex]\[ x = -0.5 \][/tex]
[tex]\[ x = -3 \][/tex]
Thus, the correct answer is:
[tex]\[ \boxed{B} \, x = -\frac{1}{2}, \, x = -3 \][/tex]
