Помогите срочно нужно Решите уравнения: 1). X^2-3x-4=0 2). 4x^2+4x+1=0 3). 5^2x =25 4)....

Решите задачу:

$x^2-3x-4=0 \\ D=b^2-4ac=(-3)^2-4*1*(-4)=9+16=25 \\ x_{1} = \frac{-b+ \sqrt{D} }{2a} = \frac{3+5}{2} = \frac{8}{2} =4 \\ \\ x_{2} = \frac{-b- \sqrt{D} }{2a} = \frac{3-5}{2} = \frac{-2}{2} =-1 \\ \\ (x-4)(x+1)=0$

$4x^2+4x+1=0 \\ D=b^2-4ac=4^2-4*4*1=16-16=0 \\ x_{1,2} = \frac{-b}{2a} = \frac{-4}{8} =- \frac{1}{2} \\ \\ (2x+1)(2x+1)=0$

$5^{2x} =25 \\ 5^{2x} =5^2 \\ 2x=2 \\ x=1$

$3^{5x+1} = 27 \\ 3^{5x+1} =3^3 \\5x+1=3 \\ 5x=2 \\ x= \frac{2}{5}$

$3x^4+5x^2-8=0 \\ x^2=t \\ 3t^2+5t-8=0 \\ D=b^2-4ac=5^2-4*3*(-8)=25+96=121 \\ t_{1} = \frac{-b+ \sqrt{D} }{2a} = \frac{-5+11}{6} = \frac{6}{6} =1 \\ \\ t_{2} = \frac{-b- \sqrt{D} }{2a} = \frac{-5-11}{6} = \frac{-16}{6} =- \frac{8}{3} =-2 \frac{2}{3} \\ \\ x^2=1 \\ x_{1} =1,x_{2} =-1$