How do you find all zeros of the function $f(x)=4x^2+29x+30$ ?

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Answer 1 · 2016-03-30T22:49:30

$x_1=-5/4$ and $x_2=-6$

From the given equation $f(x) = 4x^2+29x+30$

Set $f(x)=0$

$4x^2+29x+30=0$

Try factoring

$(4x+5)(x+6)=0$

Set both factors to zero to solve for the roots

$4x+5=0$

$4x=-5$

$x=-5/4$ this is a zero

the other one

$x+6=0$

$x=-6$
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We can use another method. Using the Quadratic Formula

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

From $4x^2+29x+30=0$ , let $a=4$ , $b=29$ , $c=30$

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

$x=(-29+-sqrt(29^2-4(4)(30)))/(2(4))$

$x=(-29+-sqrt(841-480))/(8)$

$x=(-29+-sqrt(361))/(8)$

$x=(-29+-19)/(8)$

$x_1=(-10)/(8)=-5/4$

$x_2=(-48)/(8)=-6$

God bless....I hope the explanation is useful

How do you find all zeros of the function f(x)=4x^2+29x+30f(x)=4x^2+29x+30?