Answer:[tex]\frac{7(x-3)}{4(x+5)}=\frac{3(x-5)}{5(x+4)}[/tex]
Doing criss cross multiplication
7(x-3)*5(x+4)=3(x-5)*4(x+5)
35(x*x+4*x-3*x-3*5)=20(x²-25)
35(x²+4x-3x-15)=20(x²-25)
35(x²+x-15)=20(x²-25)
35/20*(x²+x-15)=x²-25
7/4*(x²+x-15)=x²-25
7x²+7x-7*15=4x²-25*4
7x²-7x-105=4x²-100
7x²-4x²-7x-105+100=0
3x²-7x-5=0
Comparing above equation with ax²+bx+c=0
we get
a=3
b=-7
c=-5
now
By using quadratic equation formula
x=[tex]\frac{ -b±\sqrt{b²-4a}}{2a}[/tex]
Substitute value
x=[tex]\frac{ 7±\sqrt{-7²-4*3}}{2*3}[/tex]
x=[tex]\frac{ 7±\sqrt{61}}{6}[/tex]
taking positive
x=[tex]\frac{ 7+\sqrt{61}}{6}[/tex]
taking negative
x=[tex]\frac{ 7-\sqrt{61}}{6}[/tex]
