Question

4(2x-5)^2 - (3x+1)^2 =0 aflati-l pe x

Apropo ^ inseamna la putere

Answer 1

                 
$4(2x-5)^{2} - (3x+1)^{2} =0 \\ 4(4 x^{2} -20x+25)-(9 x^{2} +6x+1)=0 \\ 16 x^{2} -80x+100-9x^{2} -6x-1=0 \\ 7 x^{2} -86x+99=0 \\ \\ x_{12}= \frac{86\pm \sqrt{86^{2}+4*7*99} }{2*7}=\frac{86\pm \sqrt{7396+2772} }{14}=\frac{86\pm \sqrt{10168} }{14}= \\ =\frac{86\pm 2\sqrt{2542} }{14}=\frac{43\pm \sqrt{2542} }{7} \\ \\ x_1=\frac{43+ \sqrt{2542} }{7} \\ \\ x_2=\frac{43- \sqrt{2542} }{7}$

Answer 2

                 
$4(2x-5)^{2} - (3x+1)^{2} =0 \\ 4(4 x^{2} -20x+25)-(9 x^{2} +6x+1)=0 \\ 16 x^{2} -80x+100-9x^{2} -6x-1=0 \\ 7 x^{2} -86x+99=0 \\ \\ x_{12}= \frac{86\pm \sqrt{86^{2}+4*7*99} }{2*7}=\frac{86\pm \sqrt{7396+2772} }{14}=\frac{86\pm \sqrt{10168} }{14}= \\ =\frac{86\pm 2\sqrt{2542} }{14}=\frac{43\pm \sqrt{2542} }{7} \\ \\ x_1=\frac{43+ \sqrt{2542} }{7} \\ \\ x_2=\frac{43- \sqrt{2542} }{7}$