4x^2 - 16(p-q)x + 15p^2 - 34pq + 15q^2 = 0

Answer :

We have equation , As : 4x²  -  16 ( p - q ) x  +  15p² -  34pq + 15q² =  0

We know quadratic formula , As  :

x = $\frac{- b \pm \sqrt{b^2 - 4ac }}{2a}$  , Here a  =  4  b  =  -16 ( p - q )  and c  =  15p² -  34pq + 15q² 
SO,

$$\begin{gathered} x = \frac{- [ -16 ( p - q ) ] \pm \sqrt{[ -16 ( p - q ) {]}^2 - 4\times 4 \times \left( 15p^2 - 34pq + 15q^2 \right) }}{2 \times 4} \\ x = \frac{16 ( p - q ) \pm \sqrt{ 256 ( p^2 + q^2 - 2pq) - 16 \left( 15p^2 - 34pq + 15q^2 \right) }}{8} \\ x = \frac{16 ( p - q ) \pm \sqrt{ 256 p^2 +256 q^2 -512pq - 240 p^2 + 544pq - 240q^2 }}{8} \\ x = \frac{16 ( p - q ) \pm \sqrt{ 16 p^2 +16 q^2 +32pq }}{8} \\ x = \frac{16 ( p - q ) \pm \sqrt{ {\left(4 p + 4q \right)}^2 }}{8} \\ x = \frac{16 ( p - q ) \pm \left(4 p + 4q \right) }{8} \\ So, \\ x = \frac{16 ( p - q ) + \left(4 p + 4q \right) }{8} = \frac{16 p -16 q + 4 p + 4q }{8} = \frac{20 p -12 q }{8} = \frac{4 \left( 5 p -3q\right) }{8} \mathbf{=}\mathbf{ }\frac{\mathbf{ }\mathbf{ }\mathbf{5}\mathbf{ }\mathbf{ }\mathbf{p}\mathbf{ }\mathbf{-}\mathbf{3}\mathbf{q}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }}{\mathbf{2}} \\ And \\ x = \frac{16 ( p - q ) - \left(4 p + 4q \right) }{8} = \frac{16 p -16 q - 4 p - 4q }{8} = \frac{12 p -20 q }{8} = \frac{4 \left( 3 p -5 q\right) }{8} \mathbf{=}\mathbf{ }\frac{\mathbf{ }\mathbf{ }\mathbf{3}\mathbf{ }\mathbf{ }\mathbf{p}\mathbf{ }\mathbf{-}\mathbf{5}\mathbf{ }\mathbf{q}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }}{\mathbf{2}}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{(}\mathbf{ }\mathit{A}\mathit{n}\mathit{s}\mathbf{ }\mathbf{ }\mathbf{)} \end{gathered}$$