To determine the number and type of solutions for the given quadratic equation [tex]\( 3x^2 + 4x - 5 = 3x + 2 \)[/tex], we first need to get it into the standard form [tex]\( Ax^2 + Bx + C = 0 \)[/tex].
1. Start by isolating all terms on one side of the equation:
[tex]\[
3x^2 + 4x - 5 - (3x + 2) = 0
\][/tex]
2. Simplify the equation by combining like terms:
[tex]\[
3x^2 + 4x - 5 - 3x - 2 = 0
\][/tex]
[tex]\[
3x^2 + (4x - 3x) + (-5 - 2) = 0
\][/tex]
[tex]\[
3x^2 + x - 7 = 0
\][/tex]
Now we have the quadratic equation in the standard form:
[tex]\[
3x^2 + x - 7 = 0
\][/tex]
To determine the number and type of solutions, we need to calculate the discriminant [tex]\(\Delta\)[/tex], which is given by:
[tex]\[
\Delta = B^2 - 4AC
\][/tex]
For the quadratic equation [tex]\(Ax^2 + Bx + C = 0\)[/tex], we identify:
[tex]\[
A = 3, \quad B = 1, \quad C = -7
\][/tex]
Plug these values into the discriminant formula:
[tex]\[
\Delta = 1^2 - 4 \cdot 3 \cdot (-7)
\][/tex]
[tex]\[
\Delta = 1 - (-84)
\][/tex]
[tex]\[
\Delta = 1 + 84
\][/tex]
[tex]\[
\Delta = 85
\][/tex]
The discriminant [tex]\(\Delta\)[/tex] is 85. Based on the value of the discriminant, the type of solutions can be determined as follows:
- If [tex]\(\Delta > 0\)[/tex]: There are two distinct real solutions.
- If [tex]\(\Delta = 0\)[/tex]: There is exactly one real solution.
- If [tex]\(\Delta < 0\)[/tex]: There are two complex solutions.
Since [tex]\(\Delta = 85 > 0\)[/tex], there are two real solutions to the equation.
Therefore, the number and type of solutions for the equation [tex]\( 3x^2 + 4x - 5 = 3x + 2 \)[/tex] is:
[tex]\[
2 \text{ real solutions}
\][/tex]
