This is the concept of quadratic equations, solving the equation 2x^2-12x+20=0 using the formula we get:
x=[-b+/-sqrt(b^2-4ac)]/(2a)
from our equation, a=2,b=-12 and c=20
x=[-(-12+/-sqrt((-12)^2-4*2*20)]/(2*2)
x=[12+/-sqrt(144-160]/4
x=[12+/-sqrt(-16)]/4
x=[12+/-4i]/4
x=[3+\-i]
the answer is A] 3+/-i
It is x2-4=0 and 4x 2 = 16
Hope this helps you!!!
His brother is 2 years old
Answer:
Step-by-step explanation:
Given
m∠COD=8x+13,
m∠BOC=3x-10, and
m∠BOD=12x-6
The following addition property is true;
<BOD = <BOC + <COD
Substitute
12x-6 = 3x-10+8x+13
12x-6 = 11x+3
12x-11x = 3+6
x = 9
Get <BOD
<BOD = 12x-6
<BOD = 12(9)-6
<BOD = 108-6
<BOD = 102°
Answer:
x1=-4/5, x2=18/5 and x3=7/5
Step-by-step explanation:
![\left[\begin{array}{ccc|c}1&0&-3&-5\\3&1&2&4\\2&2&1&7\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%260%26-3%26-5%5C%5C3%261%262%264%5C%5C2%262%261%267%5Cend%7Barray%7D%5Cright%5D)
you can do linear combination between the rows:
2nd row=R2-3R1 and 3th row=R3-2R1
![\left[\begin{array}{ccc|c}1&0&-3&-5\\0&1&11&19\\0&2&7&17\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%260%26-3%26-5%5C%5C0%261%2611%2619%5C%5C0%262%267%2617%5Cend%7Barray%7D%5Cright%5D)
3th row=(3R2-R3)/15
![\left[\begin{array}{ccc|c}1&0&-3&-5\\0&1&11&19\\0&0&1&\frac{7}{5} \end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%260%26-3%26-5%5C%5C0%261%2611%2619%5C%5C0%260%261%26%5Cfrac%7B7%7D%7B5%7D%20%5Cend%7Barray%7D%5Cright%5D)
1st row=R1+3R3 and R2-11R3
![\left[\begin{array}{ccc|c}1&0&0&-4/5 \\0&1&0&18/5 \\0&0&1&7/5\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%260%260%26-4%2F5%20%5C%5C0%261%260%2618%2F5%20%5C%5C0%260%261%267%2F5%5Cend%7Barray%7D%5Cright%5D)
x1=-4/5, x2=18/5 and x3=7/5
