Question

Уравнение и его корни

Answer 1

$\frac{(3x+12)^2}{9} = 16x+1 |*9 \\ (3x+12)^2=144x+9\\ 9x^2+72x+144-144x-9=0 \\9x^2-72x+135=0 |:9 \\x^2-8x+15=0 \\D=(-8)^2-4*15=64-60=4, \sqrt{4} = 2 \\ x_{1} = 8+2/2=10/2=5 \\x_2=8-2/2=6/2=3$