3) \( a^{2}+14 a-51=0 \)

To solve the quadratic equation \( a^{2}+14 a-51=0 \), you can use the quadratic formula: \( a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here, \( a = 1 \), \( b = 14 \), and \( c = -51 \). Plugging in these values, we find the discriminant \( b^2 - 4ac = 14^2 - 4 \cdot 1 \cdot (-51) = 196 + 204 = 400 \). This means our square root is 20, and the solutions are \( a = \frac{-14 \pm 20}{2} \). So we get two possible solutions: \( a = 3 \) and \( a = -17 \). Voila! To double-check, replace \( a \) back in the original equation. For \( a = 3\): \( 3^2 + 14×3 - 51 = 0 \), and for \( a = -17\): \( (-17)^2 + 14×(-17) - 51 = 0 \). Both solutions satisfy the equation perfectly. 🎉