Question

How do you solve `4x^2-3x-7=0` using the quadratic formula?

Answer 1

`x=7/4`,x=-1

Explanation:

a = 4, b = -3, c = -7
D = `b^2-4ac`
= `(-3)^2-4*4*(-7)`
= `9+112`
=`121`
`sqrt(D)=11`
`x=(-b+-sqrt(D))/(2a)`
= `(-(-3)+-11)/(2*4)`
= `(3+-11)/8`
= `(3+11)/8,(3-11)/8`
= `14/8,-8/8`
= `7/4,-1`

Answer 2

Solution: `x =-1 , x= 1.75`

Explanation:

`4 x^2 -3 x-7=0`

Comparing with standard quadratic equation `ax^2+bx+c=0`

`a= 4 ,b=-3 ,c=-7`. Discriminant `D= b^2-4 a c` or

`D=9+112 =121` ,discriminant positive, we get two real

solutions. Quadratic formula: `x= (-b+-sqrtD)/(2a)`or

`x= (3+-sqrt 121)/8 or x = (3 +- 11)/8` or

`x = 14/8=1.75 , x = -8/8= -1`

Solution: `x =-1 , x= 1.75` [Ans]