Find the value of p for …

The given quadratic equation is: 
(2p + 1)x² - (7p + 2)x + 7p - 3 = 0
Here, a = (2p + 1), b = -(7p + 2) and c = 7p - 3
We know that, D = b² - 4ac
= [-(7p + 2)]² - 4 {tex}\times{/tex} (2p + 1) {tex}\times{/tex} (7p - 3)
= 49p² + 4 + 28p - 4(14p² - 6p + 7p - 3)
= 49p² + 4 + 28p - 4(14p² + p - 3)
= 49p² + 4 + 28p - 56p² - 4p + 12
= -7p² + 24p + 16
Since it is given that the given equation has real and equal roots, so D  = 0 i.e.,
{tex}\Rightarrow{/tex} -7p² + 24p + 16 = 0
{tex}\Rightarrow{/tex} 7p² - 24p - 16 = 0
{tex}\Rightarrow{/tex} 7p² + 4p - 28p - 16 = 0
{tex}\Rightarrow{/tex} p(7p + 4) - 4(7p + 4) = 0
{tex}\Rightarrow{/tex} (p - 4)(7p + 4) = 0
{tex}\Rightarrow{/tex} p - 4 = 0 or 7p + 4 = 0
{tex}\Rightarrow{/tex} p = 4 or {tex}p = - \frac{4}{7}{/tex}
Therefore, after substituting the value of p in the given equation, the two equations will be:
9x² - 30x + 25 = 0 [For p = 4]
{tex}\Rightarrow{/tex} 9x² - 15x - 15x + 25 = 0
{tex}\Rightarrow{/tex} 3x (3x - 5) - 5x (3x - 5) = 0
{tex}\Rightarrow{/tex} (3x - 5)² = 0
{tex}x = \frac{5}{3}{/tex}
or 
{tex}\left( {2 \times \left( { - \frac{4}{7}} \right) + 1} \right){x^2} - \left( {7\left( { - \frac{4}{7}} \right) + 2} \right)x{/tex} {tex} + 7\left( { - \frac{4}{7}} \right) - 3 = 0{/tex} [For {tex}p = - \frac{4}{7}{/tex}]
On solving we get,
x² - 14x + 49 = 0
{tex}\Rightarrow{/tex} x² - 7x - 7x + 49 = 0
{tex}\Rightarrow{/tex} x (x - 7) - 7 (x - 7)= 0
{tex}\Rightarrow{/tex} (x - 7)² = 0
{tex}\Rightarrow{/tex} x = 7
Thus, {tex}x = \frac{5}{3}{/tex} or x = 7