How do you find all zeros of f(x)=1/2x^2+5/2x-3/2?

Jul 30, 2017

$x = - \frac{5}{2} \pm \frac{1}{2} \sqrt{37}$

Explanation:

$\text{take out common factor of } \frac{1}{2}$

$\Rightarrow f \left(x\right) = \frac{1}{2} \left({x}^{2} + 5 x - 3\right) = 0$

$\text{solve "x^2+5x-3" using the "color(blue)"quadratic formula}$

$\text{with } a = 1 , b = 5 , c = - 3$

$\Rightarrow x = \frac{- 5 \pm \sqrt{25 + 12}}{2} = \frac{- 5 \pm \sqrt{37}}{2}$

$\Rightarrow x = - \frac{5}{2} \pm \frac{1}{2} \sqrt{37} \leftarrow \textcolor{red}{\text{ exact solutions}}$

$x \approx - 5.54 \text{ or " x~~ 0.54" to 2 dec/ places}$
graph{1/2x^2+5/2x-3/2 [-10, 10, -5, 5]}