Question

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

Answer 1

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

Answer 2

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