To solve the quadratic equation [tex]\( 2x^2 + 6x = 4 \)[/tex], we need to follow these steps:
1. Rearrange the equation to standard form:
The standard form of a quadratic equation is [tex]\( ax^2 + bx + c = 0 \)[/tex]. Start by moving all the terms to one side of the equation:
[tex]\[
2x^2 + 6x - 4 = 0
\][/tex]
2. Identify the coefficients:
From the equation [tex]\( 2x^2 + 6x - 4 = 0 \)[/tex], we observe that:
[tex]\( a = 2 \)[/tex],
[tex]\( b = 6 \)[/tex], and
[tex]\( c = -4 \)[/tex].
3. Use the quadratic formula:
The quadratic formula is given by:
[tex]\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
Substitute [tex]\( a = 2 \)[/tex], [tex]\( b = 6 \)[/tex], and [tex]\( c = -4 \)[/tex] into the formula:
[tex]\[
x = \frac{-6 \pm \sqrt{6^2 - 4 \cdot 2 \cdot (-4)}}{2 \cdot 2}
\][/tex]
[tex]\[
x = \frac{-6 \pm \sqrt{36 + 32}}{4}
\][/tex]
[tex]\[
x = \frac{-6 \pm \sqrt{68}}{4}
\][/tex]
[tex]\[
x = \frac{-6 \pm \sqrt{4 \cdot 17}}{4}
\][/tex]
[tex]\[
x = \frac{-6 \pm 2\sqrt{17}}{4}
\][/tex]
Simplify the expression:
[tex]\[
x = \frac{-3 \pm \sqrt{17}}{2}
\][/tex]
4. Calculate the solutions:
Now, we calculate the two possible solutions:
[tex]\[
x_1 = \frac{-3 + \sqrt{17}}{2}
\][/tex]
[tex]\[
x_2 = \frac{-3 - \sqrt{17}}{2}
\][/tex]
Evaluating these expressions, we find:
[tex]\[
x_1 \approx 0.56 \quad \text{(rounded to two decimal places)}
\][/tex]
[tex]\[
x_2 \approx -3.56 \quad \text{(rounded to two decimal places)}
\][/tex]
5. State the solutions:
Therefore, the solutions to the quadratic equation [tex]\( 2x^2 + 6x = 4 \)[/tex] are:
[tex]\[
x = 0.56 \quad \text{and} \quad x = -3.56
\][/tex]
Thus, the correct choice from the given options is:
[tex]\[x=0.56 \text{ and } x=-3.56\][/tex]
