If -5 is a root of …

Given that -5 is the root of {tex}2 x^{2}+p x-15=0{/tex}
Put x = -5 in {tex}2 x^{2}+p x-15=0{/tex} 
{tex}\Rightarrow{/tex} {tex}2(-5)^{2}+p(-5)-15=0{/tex}
{tex}\Rightarrow{/tex} {tex}50-5 p-15=0{/tex}
{tex}\Rightarrow{/tex} {tex}35 - 5p = 0 {/tex}
{tex}\Rightarrow{/tex} {tex}5p = 35 {/tex}
{tex}\therefore{/tex} {tex}p = 7{/tex}
Hence the quadratic equation p {tex}(x^2 + x) + k = 0{/tex} becomes, {tex}7\left(x^{2}+x\right)+k=0{/tex}
{tex}7 x^{2}+7 x+k=0{/tex} 
Here {tex}a = 7,\ b = 7\ and\ c = k{/tex} Given that this quadratic equation has equal roots 
{tex}\therefore{/tex} {tex}b^{2}-4 a c=0{/tex} 
{tex}\Rightarrow{/tex} {tex}7^{2}-4(7)(\mathrm{k})=0{/tex} 
{tex}\Rightarrow{/tex} {tex}49 - 28k = 0 {/tex}
{tex}\Rightarrow{/tex} {tex}49 = 28k {/tex}
{tex}\therefore{/tex} k =  {tex}\frac{49} {28}{/tex} = {tex}\frac{7} {4}{/tex}