Question

solve for 16^2+(X+7)^2=(22-2x/5)^2

Answer 1

16^2+(X+7)^2=((22-2x)/5)^2
256+(x+7)^2=((22-2x)/5)^2
256+(x+7)^2=((22-2x)/5)((22-2x)/5)
256+(x+7)^2=((22-2x)(22-2x))/25
256+(x+7)^2=484-48x-48x+4x^2/25
256+(x+7)^2=484-96x+4x^2/25
256+(x+7)(x+7)=484-96x+4x^2/25
256+(x^2+14x+49)=484-96x+4x^2/25
256+x^2+14x+49=484-96x+4x^2/25
x^2+14x+305=484-96x+4x^2/25
25(x^2+14x+305)=484-96x+4x^2
(25x^2+350x+7625)=484-96x+4x^2
(21x^2+350x+7625)=484-96x
(21x^2+446x+7625)=484
21x^2+446x+7141=0

Ax^2+Bx+C
((-b-sqrt(b^2-4ac)/(2a)
((-b+sqrt(b^2-4ac)/(2a)

answer DNE