The problem corrected is the following
4j^2+3j-28=0
In a quadratic equation
q(x) = ax^2 + bx + c
The discriminant is = b^2 - 4ac
If b^2 - 4ac > 0, then the roots are real.
If b^2 - 4ac < 0 then the roots are imaginary
In this problem
b^2 - 4ac = 3^2 – 4(4)(-28)=457
<span>457>0 </span>then the two roots must be real
<span>For the quadratic equation there are two real solutions</span>
Answer:
90 in
Step-by-step explanation:
15 x 6 = 90 in
This is because
6 x 12 hrs = 3 days