How do you solve using the quadratic formula method $4 x^{2} + 2 = 7 x$ $4x^2 + 2 = 7x$ ?

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Answer 1 · 2016-05-21T11:42:51

The solutions for the quadratic equation are:

$x = \frac{7 + \sqrt{17}}{8}$ $color(blue)(x =(7+sqrt(17))/8$

$x = \frac{7 - \sqrt{17}}{8}$ $color(blue)(x =(7-sqrt(17))/8$

$4 x^{2} + 2 = 7 x$ $4x^2 + 2 = 7x$

$4 x^{2} - 7 x + 2 = 0$ $4x^2 - 7x + 2 = 0$

The equation is of the form $a x^{2} + b x + c = 0$ $color(blue)(ax^2+bx+c=0$ where:
$a = 4, b = - 7, c = 2$ $a=4, b=-7, c=2$

The Discriminant is given by:

$Delta=b^2-4*a*c$

$= (-7)^2-(4* 4 * 2)$

$= 49 - 32 = 17$

The solutions are found using the formula
$x=(-b+-sqrtDelta)/(2*a)$

$x = (-(-7)+-sqrt(17))/(2*4) = (7+-sqrt(17))/8$

$x =(7+sqrt(17))/8$

$x =(7-sqrt(17))/8$

How do you solve using the quadratic formula method 4x^2 + 2 = 7x4x2+2=7x4x^2 + 2 = 7x?