Question 268688: -4j^2+3j-28=0
Answer by persian52(161)   (Show Source):
You can put this solution on YOUR website! You want to solve using the quadratic formula? IF yes then here!
-4j^(2)+3j-28=0
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Since -28 does not contain the variable to solve for, move it to the right-hand side of the equation by adding 28 to both sides.
-4j^(2)+3j=28
To set the left-hand side of the equation equal to 0, move all the expressions to the left-hand side.
-4j^(2)+3j-28=0
Multiply each term in the equation by -1.
-4j^(2)*-1+3j*-1-28*-1=0*-1
Simplify the left-hand side of the equation by multiplying out all the terms.
4j^(2)-3j+28=0*-1
Multiply 0 by -1 to get 0.
4j^(2)-3j+28=0
Use the quadratic formula to find the solutions. In this case, the values are a=4, b=-3, and c=28.
j=(-b\~(b^(2)-4ac))/(2a) where aj^(2)+bj+c=0
Substitute in the values of a=4, b=-3, and c=28.
j=(-(-3)\~((-3)^(2)-4(4)(28)))/(2(4))
Multiply -1 by the -3 inside the parentheses.
j=(3\~((-3)^(2)-4(4)(28)))/(2(4))
Simplify the section inside the radical.
j=(3\i~(439))/(2(4))
Simplify the denominator of the quadratic formula.
j=(3\i~(439))/(8)
First, solve the + portion of ±.
j=(3+i~(439))/(8)
Next, solve the - portion of ±.
j=(3-i~(439))/(8)
The final answer is the combination of both solutions.
j=(3+i~(439))/(8),(3-i~(439))/(8)
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