How do I use the quadratic formula to solve $f (x) = 4 x^{2} + 12 x + 9$ $f(x) = 4x^2 + 12x + 9$ ?

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Answer 1 · 2014-09-30T03:19:58

$a x^{2} + b x + c = 0$ $ax^2+bx+c=0$

$4 x^{2} + 12 x + 9 = 0$ $4x^2+12x+9=0$

$a = 4, b = 12, c = 9$ $a=4, b=12, c=9$

Quadratic formula

$x = \frac{- (12) \pm \sqrt{{(12)}^{2} - 4 (4) (9)}}{2 (4)}$ $x=(-(12)+-sqrt((12)^2-4(4)(9)))/(2(4))$

$x = \frac{- 12 \pm \sqrt{144 - 144}}{8}$ $x=(-12+-sqrt(144-144))/8$

$x = \frac{- 12 \pm \sqrt{0}}{8}$ $x=(-12+-sqrt(0))/8$

$x = \frac{- 12}{8}$ $x=(-12)/8$

$x = - \frac{3}{2}$ $x=-3/2$

$x = - 1.5$ $x=-1.5$

How do I use the quadratic formula to solve f(x) = 4x^2 + 12x + 9f(x)=4x2+12x+9f(x) = 4x^2 + 12x + 9?