Question

(x+2)^3-x(3x+1)^2+(2x+2)(4x^2-2x+1)=42

Answer 1

(x+2)^3-x(3x+1)^2+(2x+2)(4x^2-2x+1)=42

$x^{3}+6x^{2}+12x+8-9x^{3}-6x^{2}-x+8x^{3}-4x^{2}+2x+2=42 \\ 9x+10+4x^{2}=42 \\ 4x^{2}+9x-32=0$

D=b²-4ac = 9² - 4*4*(-32) = 81 + 512 = 593;
x1 = $\frac{-b+ \sqrt{D} }{2a} = \frac{-9+ \sqrt{593} }{8}$
x2 = $\frac{-b- \sqrt{D} }{2a} = \frac{-9- \sqrt{593} }{8}$