Please help me solve this algorithm: 81^x+1=9^square root of x.?

81^(x+1)=9^(√x)
3^4(x+1)=3^(2√x)

4(x+1)=2√x
(4x+4)^2=(2√x)^2
16x^2+32x+16=0=4x
16x^2+32x-4x+16=0
16x^2+28x+16=0
x^2+7/4x+1=0
x^2+7/4x=-1
x^2+7/4x+49/64=-1+49/64
(x+7/8)^2=-15/64
x+7/8=± √(-15/64)
x=-7/8±i√(15)/8
.....-7+i√(15).....-7-i√(15)
x=--------------- , ----------------- answer//
.........8..................8

What algorithm? This is an exponential equation. Notice that 81 = 9^2, so what you have here is: 9^(2x + 2) = 9^(sqr(x)). When the exponentials are equal and have the same base, the exponents, or powers, must be equal. So, therefore,

2x + 2 = sqr(x), and 4x^2 + 8x + 4 = x, and so, 4x^2 + 7x + 4 = 0. Applying the quadratic formula: x = -(7/8) +/- (i/8)(sqr(15))

81^(x + 1) = 9^(√x)
(9²)^(x + 1) = 9^(√x)
(9)^(2x + 2) = 9^(√x)

comparing index
(2x + 2) = √x
squaring both sides
4x² + 8x + 4 = x
4x² + 7x + 4 = 0
a = 4, b = 7, c = 4
x = { –7 ± √( 49 - 64 )} / 8
x = { –7 ± √( 49 - 64 )} / 8
x = { –7 ± √(– 15)} / 8
--------

81^x+1=9^square root of x
(9²)^x+1=9^√ x
(9)^2x+2=9^√ x

2x+2=√x
4x²+8x+4=x
4x²+7x+4=0
x= [-b±√ (b²-4ac)]/2a
x= [-7±√ (49-4*4*4)]/2*4
x= [-7±√ (49-64)]/8
x= [-7±√ (-15)]/8
x= [-7± i√15]/8

81^(x+1) = 9^sqrt(x)
9^(2x+2)=9^sqrt(x)

equating indicies,

2x + 2 = sqrt(x)
(2x+2)^2 = x
4x^2 + 8x + 4 = x
4x^2 + 7x + 4 = 0

now using the quadratic formula:

x = [-b+/-sqrt(b^2-4ac)]/2a
x = [-7+/-sqrt(7^2-4*4*4)]/2*4
x = (-7+/- sqrt(-15))/8

so,
x = (-7+sqrt(-15)/8
x = -0.875 + 0.484i

and

x = (-7-sqrt(-15))/8
x = -0.875 - 0.484i

9^(2x+2)=9^(\/'''x'''
4x^2+4+8x=x
4x^2+7x+4=0
x=[-7+ & -(49-64)^1/2]/8
there is not such a real x

4^x = 16 root 2 4^x = (4^2)(4^1/4) 4^x = 4^(2 + 1/4) x = (2 + 1/4) x = 9/4

81^(x + 1) = 9^(√x)
(9^2)^(x + 1) = 9^(√x)
9^[2(x + 1)] = 9^(√x)
2(x + 1) = √x
(2x + 2)^2 = x
4x^2 + 4x + 4x + 4 - x = 0
4x^2 + 7x + 4 = 0
x = [-b ±√(b^2 - 4ac)]/2a

a = 4
b = 7
c = 4

x = [-7 ±√(49 - 64)]/8
x = [-7 ±√(-15)]/8
x = [-7 ±i√15]/8 (i = imaginary number)

∴ x = [-7 ±i√15]/8