Question

find the zeros of quadratic polynomial​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

$f(x) = 6{x}^{2} - 7x - 3$

To find the zeroes:

f(x) = 0

$f(x) = 6{x}^{2} - 7x - 3 = 0\\ \implies 6{x}^{2} + 2x - 9x - 3 = 0 \\ \implies 2x(3x + 1) -3(3x + 1) = 0\\ \implies (2x - 3)(3x + 1) =0 \\ \implies 2x - 3 = 0 \; ; 3x + 1 = 0$

2x - 3 = 0

x = $\dfrac{3}{2}$

3x + 1 = 0

x = $\dfrac{-1}{3}$

Zeroes of $f(x) = 6{x}^{2} - 7x - 3$ are $\dfrac{3}{2}$ and $\dfrac{-1}{3}$