in the equation2x^2-3x-5=0?

Question

State the value of a,b, and c.
Solve the above equation using the quadratice formula.

Answer 1

ax^2 + bx + c = 0
2x^2 - 3x - 5=0

a = 2
b = -3
c = -5

"Quadratic formula":
x = [-b +/- √ (b^2 - 4ac)] / 2a
x = [3 +/- √ (-3^2 - 4(2)(-5)] / 2(2)
x = [3 +/- √ (9 - (-40)] / 4
x = [3 +/- √ (49)] / 4
x = [3 +/- 7] / 4

2 answers:
x = (3+7) / 4, and x = (3-7) / 4
x = 10/4, and x = -4/4
x = 5/2, and x = -1

Answer 2

a = 2 , b = - 3 , c = - 5
x = [ 3 ± √(9 + 40) ] / 4
x = [ 3 ± √(49) ] / 4
x = [ 3 ± 7 ] / 4
x = 10 / 4 , x = - 4 / 4
x = 5 / 2 , x = - 1

Answer 3

standard quadratic equation is:
ax^2+bx+c=0so
so a=2,b=-3 &c=-5
x={-b+- sq rt(b^2-4ac)}/2a
or x={3+-sq rt (9-4*2*-5)}/2*2
or x={3+-sq rt(49)}/4
so x= {3+-7}/4
or x=10/4 or -4/4
so x=2 1/2 or -1 ans

Answer 4

2x^2-3x-5=0
a=2
b= -3
c= -5

2x^2-3x-5=0
x= [-b+-sqrt(b^2-4ac) ] /2a

x= [3+-sqrt(9+40)] /4

x = (3+-7) /4
x = 10/4 or -1
x=5/2 or -1