Try Something Different

$\begin{aligned} \color{#3D99F6}{5x^2 + 3x} -14 & = \color{#3D99F6}{\left(\sqrt{5}x - \frac{3}{2\sqrt{5}} \right)^2 - \frac{9}{20}} - 14 \quad \quad \small \color{#3D99F6}{\text{Perfecting square}} \\ & = \color{#3D99F6}{\left(\sqrt{5}x - \frac{3}{2\sqrt{5}} \right)^2} - 14.45 \quad \quad \small \color{#3D99F6}{\text{Since }\left(\sqrt{5}x - \frac{3}{2\sqrt{5}} \right)^2 \ge 0} \\ \implies 5x^2 + 3x -14 & \ge \boxed{-14.45} \end{aligned}$

$5x^{2} + 3x -14$ can be formed as $ax^{2} +bx + c$ with $a$ = 5, $b$ = 3, and $c$ = -14.

To find the minimum value, use the formula $\frac{-D}{4a}$ with $D$ = $b^{2} -4ac$ .

$\frac{-D}{4a}$ = $\frac{-b^{2} +4ac}{4a}$ = $\frac{-3^2 + 4 \times 5 \times (-14)}{4 \times 5}$ = $\frac{-289}{20}$ = $\boxed{-14.45}$

Factor: 5x^2 + 3x -14 = (5x -7) * (x + 2)

Zeroes are 1.4 and -2 so vertex is at (1.4-2)/2 = -0.3

5(-0.3)^2 + 3(-0.3) -14 = 0.45 - 0.9 -14 = -14.45

Alternatively:

Vertex is at -b/2a = -3/(2*5) = -0.3

f(-0.3) = - 14.45

Complete the square to find this:

$f(x) = 5x^2 + 3x - 14\\ = 5 \left( x^2 + \dfrac{3}{5} x + \left( \dfrac{3}{2 \times 5} \right)^2 - \left( \dfrac{3}{2 \times 5} \right)^2 \right) - 14\\ = 5 \left( \left(x+\dfrac{3}{10} \right) ^2 - \dfrac{9}{100} \right) - 14\\ = 5 \left(x + \dfrac{3}{10} \right) ^2 - \dfrac{45}{100} - 14\\ =5 \left(x + \dfrac{3}{10} \right)^2 - 14\dfrac{45}{100}$

From here, we know that the coordinates of the minimum point is $\left( -\dfrac{3}{10}, -14\dfrac{45}{100} \right)$ .

This means that the minimum value of $f$ is $-14\dfrac{45}{100} = \boxed{-14.45}$