quadratic equation

4x^2+40x-51=0

 

a=4

b=40

c=-51

 

Just substitute these values into the quadratic equation and you will have you answers.

 

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a} \)

Solve for x:
4 x^2+40 x-51 = 0

Divide both sides by 4:
x^2+10 x-51/4 = 0

Add 51/4 to both sides:
x^2+10 x = 51/4

Add 25 to both sides:
x^2+10 x+25 = 151/4

Write the left hand side as a square:
(x+5)^2 = 151/4

Take the square root of both sides:
x+5 = sqrt(151)/2 or x+5 = -sqrt(151)/2

Subtract 5 from both sides:
x = sqrt(151)/2-5 or x+5 = -sqrt(151)/2

Subtract 5 from both sides:
Answer: |  x = sqrt(151)/2-5                 or                     x = -5-sqrt(151)/2