# How do you solve y=2x^2+5x-1 using the quadratic formula?

Mar 2, 2016

I will start you on the way so that you can complete the calculation

#### Explanation:

Given:$\text{ } y = 2 {x}^{2} + 5 x - 1$........................(1)

By solve I assume you mean solve for:$\text{ } 2 {x}^{2} + 5 x - 1 = 0$

Standard form equation: $\text{ } y = a {x}^{2} b x + c$

Where:$\text{ } x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$ ............(2)

and
$a = 2$
$b = 5$
$c = \left(- 1\right)$
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Substitution in equation (2) gives

$x = \frac{- 5 \pm \sqrt{{5}^{2} - 4 \left(2\right) \left(- 1\right)}}{2 \left(2\right)}$

$\frac{- 5 \pm \sqrt{25 + 8}}{4}$

$\textcolor{b l u e}{\text{I will let you finish this bit. There should be 2 answers}}$
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You did not ask for vertex coordinates or x-intercept or y-intercept so not included!