# How do you find the zeros of  y = 7x^2 + 5x – 3  using the quadratic formula?

Feb 4, 2016

$x = \frac{- 5 + \sqrt{109}}{14}$

$x = \frac{- 5 - \sqrt{109}}{14}$

#### Explanation:

$y = 7 {x}^{2} + 5 x - 3$ is a quadratic equation in standard form, ${x}^{2} + b x + c$, where $a = 7 , b = 5 , c = - 3$.

The zeroes of a quadratic equation are the x-intercepts where $y = 0$. To determine the zeroes, solve for $x$.

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

Substitute the known values from the quadratic equation into the quadratic formula.

$x = \frac{- 5 \pm \sqrt{{5}^{2} - \left(4 \cdot 7 \cdot - 3\right)}}{2 \cdot 7}$

Simplify.

$x = \frac{- 5 \pm \sqrt{25 - \left(- 84\right)}}{14}$

Simplify.

$x = \frac{- 5 \pm \sqrt{109}}{14}$

$109$ is a prime number.

Solve for $x$.

$x = \frac{- 5 + \sqrt{109}}{14}$

$x = \frac{- 5 - \sqrt{109}}{14}$