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

${x}_{1} = - \frac{5}{4}$ and ${x}_{2} = - 6$

Explanation:

From the given equation $f \left(x\right) = 4 {x}^{2} + 29 x + 30$

Set $f \left(x\right) = 0$

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

Try factoring

$\left(4 x + 5\right) \left(x + 6\right) = 0$

Set both factors to zero to solve for the roots

$4 x + 5 = 0$

$4 x = - 5$

$x = - \frac{5}{4}$ 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

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

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

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$x = \frac{- 29 \pm \sqrt{{29}^{2} - 4 \left(4\right) \left(30\right)}}{2 \left(4\right)}$

$x = \frac{- 29 \pm \sqrt{841 - 480}}{8}$

$x = \frac{- 29 \pm \sqrt{361}}{8}$

$x = \frac{- 29 \pm 19}{8}$

${x}_{1} = \frac{- 10}{8} = - \frac{5}{4}$

${x}_{2} = \frac{- 48}{8} = - 6$

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