How do you solve $y = 16x^2 + 40x + 25$ using the quadratic formula?

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Answer 1 · 2016-04-29T14:46:32

The solution is
$x = -5/4$

$y = 16x^2 + 40x +25$

$16x^2 + 40x +25 = 0$

The equation is of the form $color(blue)(ax^2+bx+c=0$ where:

$a=16, b=40, c=25$

The Discriminant is given by:

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

$= (40)^2-(4* 16 * 25)$

$= 1600 - 1600 = 0$

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

$x = ((-40)+-sqrt(0))/(2*16) = (-40+-0)/32$

$x = -40/32$

$x = -5/4$

How do you solve y = 16x^2 + 40x + 25y = 16x^2 + 40x + 25 using the quadratic formula?