help....3..........

4^x-3*2^(x+2)+2^5=0

$\begin{array}{rcll} 4^x-3\cdot 2^{x+2}+2^5 &=& 0 \\ 4^x-3\cdot 2^x\cdot 2^2 +2^5 &=& 0 \\ 4^x-12\cdot 2^x +2^5 &=& 0 \\ (2\cdot 2)^x-12\cdot 2^x +2^5 &=& 0 \\ 2^x \cdot 2^x-12\cdot 2^x +2^5 &=& 0 \qquad & | \qquad z=2^x\\ z^2 -12z +32 &=& 0 \end{array}$

$\boxed{~ \begin{array}{rcll} ax^2+bx+c &=& 0 \\ x &=& \dfrac{-b \pm \sqrt{b^2-4ac} }{2a} \end{array} ~}$

$\begin{array}{rcll} z^2 -12z +32 &=& 0 \quad | \quad a=1 \quad b = -12 \quad c = 32 \\ z &=& \dfrac{-(-12) \pm \sqrt{(-12)^2-4\cdot 1 \cdot 32} }{2\cdot 1} \\ z &=& \dfrac{ 12 \pm \sqrt{144-128} }{2} \\ z &=& \dfrac{ 12 \pm \sqrt{^16} }{2} \\ z &=& \dfrac{ 12 \pm 4 }{2} \\ z &=& 6 \pm 2 \\\\ z_1 &=& 6+2 \\ \mathbf{z_1} &\mathbf{=}& \mathbf{8} \\\\ z_2 &=& 6-2 \\ \mathbf{z_2} &\mathbf{=}& \mathbf{4} \end{array}$

$\begin{array}{rcll} z &=& 2^x \qquad & | \qquad \ln{()} \\ \ln{(z)} &=& x \ln{(2)} \\ \ln{(z)} &=& x \ln{(2)} \\ x &=& \frac{ \ln{(z)} }{ \ln{(2)} } \\\\ x_1 &=& \frac{ \ln{(z_1)} }{ \ln{(2)} } \\ x_1 &=& \frac{ \ln{(8)} }{ \ln{(2)} } \\ x_1 &=& \frac{ \ln{(2^3)} }{ \ln{(2)} } \\ x_1 &=& \frac{ 3 \ln{(2)} }{ \ln{(2)} } \\ \mathbf{x_1} &\mathbf{=}& \mathbf{3} \\\\ x_2 &=& \frac{ \ln{(z_2)} }{ \ln{(2)} } \\ x_2 &=& \frac{ \ln{(4)} }{ \ln{(2)} } \\ x_2 &=& \frac{ \ln{(2^2)} }{ \ln{(2)} } \\ x_2 &=& \frac{ 2 \ln{(2)} }{ \ln{(2)} } \\ \mathbf{x_2} &\mathbf{=}& \mathbf{2} \\\\ \end{array}$