How do you find the zeros, real and imaginary, of $y=-2x^2+x+3$ using the quadratic formula?

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Answer 1 · 2015-11-27T13:50:39

$x = +1 1/2 color(white)(.........)x=-1$

${EEx : AA x in RR}color(white)(....)$ Calculating y is just a matter of substituting for $x$ . However it is obvious that $y=0$

Given: $color(white)(..)y=-2x^2+x+3$

Standard form equation: $color(white)(..)y=ax^2+bx+c$

$color(white)(.....) x= (-b+-sqrt(b^2-4ac))/(2a)$

In this case:

$a=(-2)$
$b=1$
$c=3$

So $color(white)(....) x= (-1+-sqrt(1^2-4(-2)(3)))/(2(-2))$

$color(white)(....) x= (-1+-sqrt(1+24))/(-4)$

$x=(-1+-5)/(-4)$

$x = +1 1/2 color(white)(.........)x=-1$
Tont B

How do you find the zeros, real and imaginary, of y=-2x^2+x+3 y=-2x^2+x+3 using the quadratic formula?