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

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

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Answer 1 · 2016-03-02T15:56:43

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

Explanation:

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

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

Standard form equation: $y = a x^{2} b x + c$ $" "y=ax^2bx+c$

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

and
$a = 2$ $a=2$
$b = 5$ $b=5$
$c = (- 1)$ $c=(-1)$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Substitution in equation (2) gives

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

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

$I will let you finish this bit. There should be 2 answers$ $color(blue)("I will let you finish this bit. There should be 2 answers")$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

You did not ask for vertex coordinates or x-intercept or y-intercept so not included!

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

1 Answer

Explanation:

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