How do you find the zeros, real and imaginary, of $y=x^2-3x+298$ using the quadratic formula?

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Answer 1 · 2016-03-09T20:58:19

$x=3/2+ (13sqrt(7))/2 i$

$x=3/2- (13sqrt(7))/2 i$

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

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

In your case:
$a=1$
$b=-3$
$c=298$

$=> x=(-(-3)+-sqrt((-3)^2-4(1)(298)))/(2(1))$

$=> x=(3+-sqrt(-1183))/2$

'~~~~~~~~~~~~~~~~~~~~~
$169xx7=1183$
$169=13^2$
'~~~~~~~~~~~~~~~~~~~
$=> x=(3+-sqrt(-1xx7xx13^2))/2$

$=> x=(3+-13sqrt(-1)sqrt7)/2$

$x=3/2+ (13sqrt(7))/2 i$

$x=3/2- (13sqrt(7))/2 i$

How do you find the zeros, real and imaginary, of y=x^2-3x+298y=x^2-3x+298 using the quadratic formula?