Solve this with the explanation please

How many real and imaginary solutions does the equation x^2-3x=-2x-7 have?

Prove (a) cosh2(x) − sinh2(x) = 1 and (b) 1 − tanh 2(x) = sech 2(x). solution (a) cosh2(x) − sinh2(x) = ex + e−x 2 2 − 2 = e2x + 2 + e−2x 4 − = 4 = . (b) we start with the identity proved in part (a): cosh2(x) − sinh2(x) = 1. if we divide both sides by cosh2(x), we get 1 − sinh2(x) cosh2(x) = 1 or 1 − tanh 2(x) = .

For [tex]f(x) = 4x + 1[/tex] and (x) = [tex]g(x)= x^{2} -5,[/tex] find [tex](\frac{g}{f}) (x)[/tex]a. [tex]\frac{x^{2} - 5 }{4x +1 },x[/tex] ≠ [tex]-\frac{1}{4}[/tex]b. x[tex]\frac{4 x +1 }{x^{2} - 5}, x[/tex] ≠ ± [tex]\sqrt[]{5}[/tex]c. [tex]\frac{4x +1}{x^{2} -5}[/tex]d.[tex]\frac{x^{2} -5 }{4x + 1}[/tex]