To solve the quadratic equation \( 2x^2 + 3x - 5 = 0 \), we can fill in the blanks as follows:
The standard form of a quadratic equation is: \[ ax^2 + bx + c = 0 \] For the equation \( 2x^2 + 3x - 5 = 0 \):
- \( a = 2 \)
- \( b = 3 \)
- \( c = -5 \)
Now, we apply the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Filling in the blanks:
- \( -b = -3 \)
- \( b^2 - 4ac = 3^2 - 4(2)(-5) = 9 + 40 = 49 \)
- \( 2a = 2(2) = 4 \)
Thus, we can express the previous formula in filled blanks: \[ x = \frac{-3 \pm \sqrt{49}}{4} \]
Now we simplify:
- \( \sqrt{49} = 7 \)
So we have: \[ x = \frac{-3 \pm 7}{4} \]
Now we can calculate the two possible solutions:
- \( x = \frac{-3 + 7}{4} = \frac{4}{4} = 1 \)
- \( x = \frac{-3 - 7}{4} = \frac{-10}{4} = -\frac{5}{2} \)
Thus, the final solutions are: \[ x = 1 \quad \text{and} \quad x = -\frac{5}{2} \]
Summarizing the entire solution, we have:
- \( a = 2 \)
- \( b = 3 \)
- \( c = -5 \)
- \( -b = -3 \)
- \( b^2 - 4ac = 49 \)
- \( 2a = 4 \)
- \( x = \frac{-3 \pm 7}{4} \)
- Solutions: \( x = 1 \) and \( x = -\frac{5}{2} \)