How do you solve $8x^2-38x=4$ using the quadratic formula?

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How do you solve $8x^2-38x=4$ using the quadratic formula?

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Answer 1 · 2017-06-17T06:23:29

$x=-0.1025$
$x=4.85$

Explanation:

Given -

$8x^2-38x-4=0$

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

$a = 8$
$b=-38$
$c=-4$

$x=(-(38)+-sqrt(-38^2-(4xx8xx-4)))/(2xx8$

$x=(38+-sqrt(1444-(-128)))/16$

$x=(38+-sqrt(1572))/16$

$x=(38+-39.64)/16$

$x=(38-39.64)/16=(-1.64)/16=-0.1025$
$x=(38+39.64)/16=(77.64)/16=4.85$

$x=-0.1025$
$x=4.85$

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How do you solve $8x^2-38x=4$ using the quadratic formula?

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