Two pipes running together can fill …

Let time taken by tap of smaller diameter to fill the tank = x hours
Let time taken by tap of larger diameter to fill the tank = (x - 10) hours
It means that tap of smaller diameter fills {tex}{\frac{1}{x}^{th}}{/tex}part of tank in 1 hour. ................ (1)
And, tap of larger diameter fills{tex}{\frac{1}{{x - 10}}^{th}}{/tex}part of tank in 1 hour. ................ (2)
When two taps are used together, they fill tank in 75/8 hours.
In 1 hour, they fill {tex}{\frac{8}{{75}}^{th}}{/tex} part of tank {tex}\left( \frac { 1 } { \frac { 75 } { 8 } } = \frac { 8 } { 75 } \right){/tex}.................. (3)
From (1), (2) and (3),
{tex}\frac{1}{x} + \frac{1}{{x - 10}} = {\frac{8}{{75}}_ \Rightarrow }\;\frac{{x - 10 + x}}{{x(x - 10)}} = \frac{8}{{75}}{/tex}
⇒ 75(2x−10)=8(x²−10x) 
⇒ 150x - 750=8x ²−80x
⇒ 8x ²−80x−150x +750=0
⇒ 4x ²−115x +375=0 (Dividing whole equation by 2)
Comparing equation 4x ²−115x +375=0 with general equation ax ² +bx +c =0,
We get a =4, b =−115 and c =375
Applying quadratic formula {tex}x = \frac { - b \pm \sqrt { b ^ { 2 } - 4 ac } } { 2 a }{/tex} 
{tex}x = {\frac{{115 \pm \sqrt {{{( - 115)}^2} - 4(4)(375)} }}{{2 \times 4}}_ \Rightarrow }\quad x = \frac{{115 \pm \sqrt {13225 - 6000} }}{8}{/tex}
⇒ {tex}x = \frac { 15 \pm \sqrt { 725 } } { 8 }{/tex} 
⇒ {tex}x = \frac { 115 \pm 85 } { 8 }{/tex} 
⇒{tex}x = \frac { 115 + 85 } { 8 } , \frac { 115 - 85 } { 8 }{/tex}
⇒ x =25, 3.75
Time taken by larger tap = x - 10=3.75 - 10=− 6.25hours
Time cannot be in negative. Therefore, we ignore this value.
Time taken by larger tap = x - 10=25 - 10=15 hours
Therefore, time taken by larger tap is 15 hours and time is taken by smaller tap is 25 hours.