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

Answers

Answer 1

Answer: $7x^2-31x+20=0$
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:
$x-(31)/(14)=\sqrt{(401)/(196)}$

Answer 2

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

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