Question

how to solve 4n²-3n-7=0

Answer 1

`4n^2-3n-7=0\\\\\underbrace{(2n)^2-2\cdot2n\cdot(3)/(4)+\left((3)/(4)\right)^2}_((*))-\left((3)/(4)\right)^2=7\\\\(2n-(3)/(4))^2-(9)/(16)=7\\\\(2n-(3)/(4))^2=7+(9)/(16)\\\\(2n-(3)/(4))^2=(112)/(16)+(9)/(16)\\\\(2n-(3)/(4))^2=(121)/(16)\to2n-(3)/(4)=\sqrt(121)/(16)\ \vee\ 2n-(3)/(4)=-\sqrt(121)/(16)`


`2n=(11)/(4)+(3)/(4)\ \vee\ 2n=-(11)/(4)+(3)/(4)\\\\2n=(14)/(4)\ \vee\ 2n=-(8)/(4)\\\\2n=(7)/(2)\ \vee\ 2n=-2\\\\n=(7)/(4)\ \vee\ n=-1`




`(a-b)^2=a^2-2ab+b^2`

Answer 2

`4n^2-3n-7=0\\ \\ a=4, \ \ b=-3 \ \ c=-7\\ \\ \Delta = b^(2)-4ac = (-3)^(2)-4*4* (-7)= 9+112=121`


`x_(1)=(-b-√(\Delta ))/(2a) =( 3-√( 121))/(2*4)=( 3-11)/(8)= (-8 )/( 8)=-1 \\ \\x_(2)=(-b+√(\Delta ))/(2a) =( 3+√( 121))/(2*4)=( 3+11)/(8)= (14 )/( 8)= (7)/(4)=1(3)/(4)`