Question

Find x so the distance between (x,4) and (2x-4,-5) is 15

Answer 1

$(2x-4-x)^2 + (-5-4)^2 = 15^2\\ (x-4)^2 + (-9)^2 = 225\\ x^2 - 8x + 16 + 81 - 225 = 0\\ x^2 - 8x -128 = 0\\ x= \frac{8б \sqrt{(-8)^2-4*1*(-128)} }{2}=\frac{8б \sqrt{576} }{2}= \frac{8б 24 }{2}= \frac{2(4б 12) }{2}= 4б12 \\ x_1=4-12=-8 \\ x_2=4+12=16$