How do you solve $20x^2 = 13 - x$ ?

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Answer 1 · 2018-04-07T20:54:54

$x~~0.78,0.83$

Rearranging the equation
$=>20x^2+x-13=0 => (1)$
$ax^2+bx+c=0$ $=>$ standard equation

Comparing ( $1$ ) with standard equation

$a=20,b=1,c=-13$

$x=(-b+-sqrt(b^2-4ac))/(2a)$

$=>(-1+-sqrt(1^2-4(20×-13)))/(2(20))$

$=>(-1+-sqrt(1-4(-260)))/40$

$=>(-1+-sqrt(1+1040))/40$

$=>(-1+-sqrt(1041))/40$

$=>(-1+-32.26)/40$

$x~~(-1+32.26)/40,(-1-32.26)/40$

$x~~31.26/40,-33.26/40$

$x~~0.78,0.83$

How do you solve 20x^2 = 13 - x20x^2 = 13 - x?