$x\cdot (1-5x)+3=\left [ 5-2(2+5x) \right ]\cdot x-2\cdot (x+1)-\left ( x^{2}-6 \right )$
$x\cdot (1-5x)+3=\left [ 5-2(2+5x) \right ]\cdot x-2\cdot (x+1)-\left ( x^{2}-6 \right )$
Ciao,
$x\cdot (1-5x)+3=\left [ 5-2(2+5x) \right ]\cdot x-2\cdot (x+1)-\left ( x^{2}-6 \right )$
$x-5x^2+3=\left [ 5-4-10x \right ]\cdot x-2x-2- x^{2}+6$
$x-5x^2+3=\left [ 1-10x \right ]\cdot x-2x-2- x^{2}+6$
$x-5x^2+3=x-10x^2-2x-2- x^{2}+6$
$x-5x^2+3=-11x^{2}-x+4$
$x-5x^2+3+11x^{2}+x-4$
$6x^2+2x-1=0$
$\Delta = b^{2}-4ac=2^{2}-4\cdot (-1)\cdot (6)=4+24=28 $
$x_{1,2}=\frac{-b\pm \sqrt{\Delta }}{2a}=\frac{-2\pm \sqrt{28 }}{2\cdot 6}=$
$\frac{-2\pm 2\sqrt{7}}{12}=\frac{2(-1\pm \sqrt{7})}{12}=\frac{-1\pm \sqrt{7}}{6}$
$x_{1}=\frac{-1- \sqrt{7}}{6}$
$x_{2}=\frac{-1+\sqrt{7}}{6}$
saluti ?