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

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Answer 1 · 2015-05-31T12:46:15

First subtract 117 from both sides to get:

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

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

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

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

Then

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x = (-b +-sqrt(b^2-4ac))/(2a)$

$= \frac{- 3 \pm \sqrt{3^{2} - (4 \times 1 \times - 28)}}{2 \times 1}$ $=(-3+-sqrt(3^2-(4xx1xx-28)))/(2xx1)$

$= \frac{- 3 \pm \sqrt{9 + 112}}{2}$ $=(-3+-sqrt(9+112))/2$

$= \frac{- 3 \pm \sqrt{121}}{2}$ $=(-3+-sqrt(121))/2$

$= \frac{- 3 \pm \sqrt{11^{2}}}{2}$ $=(-3+-sqrt(11^2))/2$

$= \frac{- 3 \pm 11}{2}$ $=(-3+-11)/2$

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

How do you solve using the quadratic formula for 4x^2+12x+5=1174x2+12x+5=1174x^2+12x+5=117?