Question

Solve the equation for all real solutions in simplest form.
2, a, squared, plus, 4, a, minus, 6, equals, minus, 3, a
2a
2
+4a−6=
−3a

Answer 1

To solve the equation \( 2a^2 + 4a - 6 = -3a \), we first rearrange the terms to one side of the equation:

1. Add \( 3a \) to both sides:
\[
2a^2 + 4a + 3a - 6 = 0
\]
This simplifies to:
\[
2a^2 + 7a - 6 = 0
\]

2. Now, we want to solve the quadratic equation \( 2a^2 + 7a - 6 = 0 \). We can use the quadratic formula, which states that for an equation in the form \( ax^2 + bx + c = 0 \), the solutions for \( x \) are given by:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here, \( a = 2 \), \( b = 7 \), and \( c = -6 \).

3. Calculate the discriminant (\( b^2 - 4ac \)):
\[
b^2 - 4ac = 7^2 - 4 \cdot 2 \cdot (-6) = 49 + 48 = 97
\]

4. Now substitute \( a \), \( b \), and the discriminant into the quadratic formula:
\[
a = \frac{-7 \pm \sqrt{97}}{2 \cdot 2} = \frac{-7 \pm \sqrt{97}}{4}
\]

Therefore, the solutions for \( a \) are:
\[
a = \frac{-7 + \sqrt{97}}{4} \quad \text{and} \quad a = \frac{-7 - \sqrt{97}}{4}
\]

These are the two real solutions in simplest form.