Question

A quadratic equation is shown below:

Answer 1

A.
the radicand is b²-4ac
if it is in form ax²+bx+c

when it is
greater than 0, then 2 real roots
equal to 0, then 1 real root
less than 0, then no real roots

so
a=4
b=-12
c=10

b²-4ac=(-12)²-4(4)(10)=144-160<0
so 0 real roots



B.
using quadratic formula because it is easier
for
ax²+bx+c=0
$x=\frac{-b+/-\sqrt{b^2-4ac}}{2a}$

so

for 2x²-13x+21
a=2
b=-13
c=21

$x=\frac{-(-13)+/-\sqrt{(-13)^2-4(2)(21)}}{2(2)}$
$x=\frac{13+/-\sqrt{169-168}}{4}$
$x=\frac{13+/-\sqrt{1}}{4}$
$x=\frac{13+/-1}{4}$
x=(13+1)/4 or (13-1)/4
x=14/4 or 12/4
x=7/2 or 3