Question

(4+3x)^2-8=2x^2+39x срочно помагите

Answer 1

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