x²-23x +76=0 x²-9x-36=0
Let's solve both quadratic equations one by one.
1. For the equation \( x^2 - 23x + 76 = 0 \).
We can use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 1 \), \( b = -23 \), and \( c = 76 \).
Calculate the discriminant: \[ b^2 - 4ac = (-23)^2 - 4 \cdot 1 \cdot 76 = 529 - 304 = 225 \]
Calculate the roots: \[ x = \frac{23 \pm \sqrt{225}}{2 \cdot 1} \] \[ = \frac{23 \pm 15}{2} \]
This gives us two solutions: \[ x_1 = \frac{23 + 15}{2} = \frac{38}{2} = 19 \] \[ x_2 = \frac{23 - 15}{2} = \frac{8}{2} = 4 \]
So, the solutions for \( x^2 - 23x + 76 = 0 \) are: \[ x = 19 \quad \text{and} \quad x = 4 \]
2. For the equation \( x^2 - 9x - 36 = 0 \).
We again use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 1 \), \( b = -9 \), and \( c = -36 \).
Calculate the discriminant: \[ b^2 - 4ac = (-9)^2 - 4 \cdot 1 \cdot (-36) = 81 + 144 = 225 \]
Calculate the roots: \[ x = \frac{9 \pm \sqrt{225}}{2 \cdot 1} \] \[ = \frac{9 \pm 15}{2} \]
This gives us two solutions: \[ x_1 = \frac{9 + 15}{2} = \frac{24}{2} = 12 \] \[ x_2 = \frac{9 - 15}{2} = \frac{-6}{2} = -3 \]
So, the solutions for \( x^2 - 9x - 36 = 0 \) are: \[ x = 12 \quad \text{and} \quad x = -3 \]
Summary of Solutions
- For \( x^2 - 23x + 76 = 0 \): \( x = 19 \) and \( x = 4 \)
- For \( x^2 - 9x - 36 = 0 \): \( x = 12 \) and \( x = -3 \)
