How do you solve $y=2x^2+5x-1$ 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

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

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

Standard form equation: $" "y=ax^2bx+c$

Where: $" "x=(-b+-sqrt(b^2-4ac))/(2a)$ ............(2)

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

$x=(-5+-sqrt(5^2-4(2)(-1)))/(2(2))$

$(-5+-sqrt(25+8))/(4)$

$color(blue)("I will let you finish this bit. There should be 2 answers")$
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You did not ask for vertex coordinates or x-intercept or y-intercept so not included!

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