How do you solve ${{x}^{2}}+5x-4$ using the quadratic formula ?

Hint: The formula for the roots of the quadratic equation $a{{x}^{2}}+bx+c$ is equal to $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ so there are 2 roots of a quadratic equation. The term ${{b}^{2}}-4ac$ is known as the discriminant D . The property of the 2 roots depends on the value of D. If D is greater than 0 then the 2 roots are real and distinct. If the value of D is less than 0 then 2 roots are complex and one root is conjugate of another root. If the value of D is 0 then the roots are equal and real.

Complete step by step answer:
The given quadratic equation in the question is ${{x}^{2}}+5x-4$
If we compare ${{x}^{2}}+5x-4$ to quadratic equation $a{{x}^{2}}+bx+c$ then a is 1 , b is 5 and c is -4
The formula for roots of quadratic equation $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
If put the value of a, b, and c we will get the roots
So the roots of equation are $\dfrac{-5\pm \sqrt{{{5}^{2}}-4\times 1\times \left( -4 \right)}}{2\times 1}$
Solving further we get
 $\Rightarrow \dfrac{-5\pm \sqrt{{{5}^{2}}-4\times 1\times \left( -4 \right)}}{2\times 1}=\dfrac{-5\pm \sqrt{25-\left( -16 \right)}}{2}$
$\Rightarrow \dfrac{-5\pm \sqrt{{{5}^{2}}-4\times 1\times \left( -4 \right)}}{2\times 1}=\dfrac{-5\pm \sqrt{41}}{2}$

So the roots are $\dfrac{-5+\sqrt{41}}{2}$ and $\dfrac{-5-\sqrt{41}}{2}$

Note: We can see that the value of discriminant or D of the quadratic equation is 41, so the roots are real and different. We have seen that the property of roots are depend on the value of D but these dependents of root on D will apply when then coefficient of all the term in the quadratic equation or a, b, and c in the equation $a{{x}^{2}}+bx+c$ are real numbers.