How do you find the roots, real and imaginary, of $y= -2x^2 + 15x + 22$ using the quadratic formula?

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Answer 1 · 2016-01-18T11:13:35

at $y=0$ $color(white)(.........)x= + 8.756 " or " -1.256$

Given: $color(white)(...)color(brown)(y=-2x^2+15x+22)$

Using standard for of $y=ax^2+bx+c=0$

Where: $color(white)(....)x=(-b+-sqrt(b^2-4ac))/(2a)$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$a=-2$
$b=15$
$c=22$

Thus: $color(white)(.....)x=(-15+-sqrt(15^2-4(-2)(22)))/(2(-2))$

$x= (-15+-sqrt(225 + 176))/(-4)$

$x=(-15+-sqrt(401))/(-4)$

But 401 is a prime number so unable to break it down further

Using $sqrt(401)=20.025$ to 3 decimal places

$x= + 8.756 " or " -1.256$
Tony B

How do you find the roots, real and imaginary, of y= -2x^2 + 15x + 22 y= -2x^2 + 15x + 22 using the quadratic formula?