Question

a) x² + 3x - 130 = 0;
Б) x² - 7x - 120 = 0;

Answer 1

а)

${x}^{2} + 3x - 130 = 0 \\ a = 1 \\ b =3 \\ c = - 130\\ D = {b}^{2} - 4ac = {3}^{2} - 4 \times 1 \times ( - 130) = 9 + 520 = 529 \\ x_{1} = \frac{ - 3 - 23}{2} = - \frac{26}{2} = - 13 \\ x_{2} = \frac{ - 3 + 23}{2} = \frac{20}{2} = 10$

б)

${x}^{2} - 7x - 120 = 0 \\ a = 1\\ b = - 7\\ c = - 120 \\ D = {b}^{2} - 4ac = ( - 7) {}^{2} - 4 \times 1 \times ( - 120) =\\= 49 + 480 = 529 \\ x_{1} = \frac{7 - 23}{2} = - \frac{16}{2} = - 8 \\ x_{2} = \frac{7 + 23}{2} = \frac{30}{2} = 15$