SOLVING EQUATION THAT ARE TRANSFORMABLE INTO QUADRATIC EQUATION 1. x(x+5) = 22. (s+6)² = 153. (t+2)² + (t-3)² = 9You can use any method po and show your solution pa tulong i really need thisthanks​

Step-by-step explanation:1. [tex]x(x + 5) = 2 \\ x {}^{2} + 5x = 2 \\ x {}^{2} + 5x - 2 = 0[/tex][tex]quadratic \: formula \: x = \frac{ - b + - \sqrt{b {}^{2} - 4ac} }{2a} [/tex][tex]x = \frac{ - 5 + - \sqrt{5 {}^{2} - 4 \times 1 \times ( - 2) } }{2 \times 1} [/tex][tex]x = \frac{ - 5 + - \sqrt{25 - 8} }{2} [/tex][tex]x = \frac{5 + - \sqrt{33} }{2} [/tex]2. [tex](s + 6 {})^{2} = 15[/tex][tex]s + 6 = + - \sqrt{15 \\} \\ s = - 6 + - \sqrt{15} [/tex]3.[tex](t + 2) {}^{2} + (t - 3 {}) {}^{2} = 9 \\ (t {}^{2} + 4t + 4) + (t {}^{2} - 6t + 9) = 9 \\ 2t {}^{2} - 2t + 13 = 9 \\ 2t {}^{2} - 2t + 4 = 0[/tex][tex]t = \: \frac{ - ( - 2) {2}^{2} \sqrt{( - 2) {}^{2} - 4 \times 2 \times 4 } }{2 \times 2 } \\ t = \frac{2 + - \sqrt{4 - 32} }{4} \\ t = \frac{2 + - \sqrt{ - 28} }{4} \\ t = 2 + - \sqrt{4 \times( - 7)} over \: 4 \\ t = \frac{ 2+ -2i \sqrt{7} }{4} \\ t = \frac{1 + - i \sqrt{7} }{2} [/tex]

Answered by danemariel | 2024-09-01