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

Sagot :

[tex]7x^2-31x+20=0\\ \\ \Delta=(-31)^2-4.7.20=961-560=401\\ \\ x=\frac{31 \pmsqrt{401}}{2*7}=\frac{31 \pm\sqrt{401}}{14}[/tex]
[tex]7x^2-31x+20=0[/tex]
First, remove the coefficient by division
So
[tex]\frac{7x^2}{7}-\frac{31x}{7}+\frac{20}{7}=0[/tex]
Then, cancel out [tex]\frac{20}{7}[/tex]
So,
[tex]x^2-\frac{31x}{7}+\frac{20}{7}-\frac{20}{7}=-\frac{20}{7}[/tex]
Then take half of the coefficient, then square it.
[tex]\frac{-\frac{31}{7}}{2}=\frac{37}{14}[/tex]
[tex](-\frac{37}{14})^2=\frac{961}{196}[/tex]
So, The equation already looks like this:
[tex]x^2-\frac{31}{7}+\frac{961}{196}=-\frac{20}{7}+\frac{961}{196}[/tex]
So,
[tex]x^2-\frac{31}{7}+\frac{961}{196}=\frac{401}{196}[/tex]
Then factor the left side of the equation:
[tex](x-\frac{31}{14})^2=\frac{401}{196}[/tex]
Then square both sides:
[tex]x-\frac{31}{14}=\sqrt{\frac{401}{196}}[/tex]