Question

(4+3x)^2-8=2x^2+39x срочно помагите

Answer 1

$(4+3x)^2-8=2x^2+39x \\ 16+24x+9x^2-8=2x^2+39x \\ 8+24x+9x^2=2x^2+39x \\ 8+23x+9x^2-2x^2-39x=0 \\ 7x^2-15x+8=0 \\ D= \sqrt{(-15)^2-4*7*8}= \sqrt{225-224} = \sqrt{1} =б1 \\ \\ x_1= \frac{15+1}{14} = \frac{16}{14} = \frac{8}{7} \\ \\ x_2= \frac{15-1}{14} =1$