This is C++

Use the bisection algorithm to find an approximate solution z to the equation x^5.3 + (3.5)^x = N where N is 1234567, z is correct to 2 significant figures and z is correct to 2 decimal places.

Use the bisection algorithm to find an approximate solution z to the equation *5.3 + (3.5)* = N where N is your 7-digit phone number, and: (a) z is correct to 2 significant figures. (b) z is correct to 2 decimal places. The output should show 2 separate results clearly labeled. The walk-through is the perfect example for the output. Walkthrough with input f(x) = x + 2*, T = 200, A = 5, B = 10, and 8 = 0.005: // Each line in the table corresponds to fixed values of A and B (and z between them). z 11 B 10 7.5 6.25 5.625 A 5 7.5 6.25 5.625 5.3125 5.3125 5.46875 5.390625 5.390625 5.4296875 5.41015625 5.400390625 5.400390625 5.4052734375 z - A f(z) 2.5 602.894... 1.25 320.249... .625 227.329... .3125 189.672... .15625 207.840... .078125 198.596... .0390625 203.177... .01953125 200.876... .009765625 199.733... .0048828125 (200.304...) 5.46875 5.4296875 5.41015625 11