Question

n^2+201n-5400 = 0
plz solve.​

Answer 1

Answer:

Step-by-step explanation:n²+201n-5400=0

or,n²+225n-24n-5400=0

or,n(n+225)-24(n+225)=0

or,(n+225)(n-24)=0

either n+225=0 or n-24=0

either n=-225 or n=24

Ans:The roots of the equation are -225 and 24 which are the values of n.

Answer 2

$\mathbb{\large{\pink{Question:-}}}$

$\mathtt{\underline{\ n^{2}+201n-5400=0}}$

$\textit{\large{\red{\underline{\underline{Step by step Explanation:-}}}}}$

$\textit{\underline{Given}}$

$\mathtt{\ a=1,,b=201,,c=-5400}$

$\textbf{\underline{By Quadratic formula,,}}$

$\mathtt{\ D=b^{2}-4ac}$

$\mathtt\implies{\ D=201^{2}-4 \times 1\times (-5400)}$

$\mathtt\implies{\ D=40401+21600}$

$\mathtt\implies{\ D=62001}$

$\mathtt\implies{D>0}$

$\textbf{\underline{Hence,,}}$

$\mathtt{n=\frac{-b\pm \sqrt{D}}{2a}}$

$\mathtt\implies{n=\frac{-201\pm \sqrt{62001}}{2\times 1}}$

$\mathtt\implies{n=\frac{-201\pm 249}{2}}$

$\mathtt{n=\frac{-201+249}{2}}$

$\mathtt\implies{n=\frac{48}{2}}$

$\mathtt\implies{n=24}$

$\textbf{\underline{And,,}}$

$\mathtt{n=\frac{-201-249}{2}}$

$\mathtt\implies{n=\frac{-500}{2}}$

$\mathtt\implies{n=-225}$

$\textbf{\underline{Hence,,}}$

$\mathtt\implies{n= 24,,and,,-225}$

$\textbf{\large{\blue{Hope it helps you..}}}$

$\textbf{\large{\red{Please mark it as Brainliest answer.. thanks}}}$