The quadratic formula used to solve the equation
is ![x=\frac{-3 \pm \sqrt{(3)^{2}-4(5)(-4)}}{2(5)}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B-3%20%5Cpm%20%5Csqrt%7B%283%29%5E%7B2%7D-4%285%29%28-4%29%7D%7D%7B2%285%29%7D)
Explanation:
The equation is ![5x^{2} +3x-4=0](https://tex.z-dn.net/?f=5x%5E%7B2%7D%20%2B3x-4%3D0)
The equation is of the form ![ax^{2} +bx+c=0](https://tex.z-dn.net/?f=ax%5E%7B2%7D%20%2Bbx%2Bc%3D0)
Thus, ![a=5, b=3,c=-4](https://tex.z-dn.net/?f=a%3D5%2C%20b%3D3%2Cc%3D-4)
To find the quadratic formula, the general formula to find the quadratic roots is
Hence, substituting the values of a,b,c in the formula, we get,
![x=\frac{-3 \pm \sqrt{(3)^{2}-4(5)(-4)}}{2(5)}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B-3%20%5Cpm%20%5Csqrt%7B%283%29%5E%7B2%7D-4%285%29%28-4%29%7D%7D%7B2%285%29%7D)
Thus, Option A is the correct answer.
The quadratic formula used to solve the equation
is ![x=\frac{-3 \pm \sqrt{(3)^{2}-4(5)(-4)}}{2(5)}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B-3%20%5Cpm%20%5Csqrt%7B%283%29%5E%7B2%7D-4%285%29%28-4%29%7D%7D%7B2%285%29%7D)