How do you solve using the quadratic formula for $4x^2+12x+5=117$ ?

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Answer 1 · 2015-05-31T12:46:15

First subtract 117 from both sides to get:

$4x^2+12x-112 = 0$

Notice all the terms are divisible by $4$ , so divide both sides by $4$ to get:

$x^2+3x-28 = 0$

This is in the form $ax^2+bx+c = 0$ with $a=1$ , $b=3$ , $c=-28$

Then

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

$=(-3+-sqrt(3^2-(4xx1xx-28)))/(2xx1)$

$=(-3+-sqrt(9+112))/2$

$=(-3+-sqrt(121))/2$

$=(-3+-sqrt(11^2))/2$

$=(-3+-11)/2$

That is $x = -7$ or $x = 4$

How do you solve using the quadratic formula for 4x^2+12x+5=1174x^2+12x+5=117?