Question

7x^2-31x+20 I Know The Answer, I Just Don't Know The Steps To Solving This Equation.

Answer 1

`7x^2-31x+20=0\\ \\ \Delta=(-31)^2-4.7.20=961-560=401\\ \\ x=\frac{31 \pmsqrt{401}}{2*7}=(31 \pm√(401))/(14)`

Answer 2

`7x^2-31x+20=0`
First, remove the coefficient by division
So

`(7x^2)/(7)-(31x)/(7)+(20)/(7)=0`
Then, cancel out
`(20)/(7)`
So,

`x^2-(31x)/(7)+(20)/(7)-(20)/(7)=-(20)/(7)`
Then take half of the coefficient, then square it.

`(-(31)/(7))/(2)=(37)/(14)`

`(-(37)/(14))^2=(961)/(196)`
So, The equation already looks like this:

`x^2-(31)/(7)+(961)/(196)=-(20)/(7)+(961)/(196)`
So,

`x^2-(31)/(7)+(961)/(196)=(401)/(196)`
Then factor the left side of the equation:

`(x-(31)/(14))^2=(401)/(196)`
Then square both sides:

`x-(31)/(14)=\sqrt{(401)/(196)}`