# How do you find the zeros, real and imaginary, of y= -7x^2-5x+3 using the quadratic formula?

Feb 21, 2018

The zeroes (or roots, as they are also called) of $y = - 7 {x}^{2} - 5 x + 3$ are $x = \frac{5 \pm \sqrt{109}}{-} 14$

#### Explanation:

The problem says to use the quadratic formula, so we will use that. The quadratic formula is $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$. $a$,$b$, and $c$ come from $a {x}^{2} + b x + c$, the standard form of a quadratic equation.

The equation is already in standard form (thank goodness). $a = - 7$, $b = - 5$, and $c = 3$.
Plugging these into the quadratic formula:
$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
$x = \frac{- \left(- 5\right) \pm \sqrt{{\left(- 5\right)}^{2} - 4 \left(- 7\right) \left(3\right)}}{2 \left(- 7\right)}$
$x = \frac{5 \pm \sqrt{25 - \left(- 28\right) \left(3\right)}}{-} 14$
$x = \frac{5 \pm \sqrt{25 - \left(- 84\right)}}{-} 14$
$x = \frac{5 \pm \sqrt{25 + 84}}{-} 14$
$x = \frac{5 \pm \sqrt{109}}{-} 14$
109 is a prime number, so $\sqrt{109}$ is in simplest form. Therefore our answer is $x = \frac{5 \pm \sqrt{109}}{-} 14$