# Quadratic Formula Example with rational roots:

*Solve the equation x^{2 }+ 3x – 4 = 0

using quadratic formula?

If we find its factors, we have (x+4)(x-1)=0, which yields -4 and 1 as the roots of equation. Since we are solving **quadratic formula examples**, let's do it with quadratic formula. Comparing it with ax^{2 }+ bx +c = 0

, we have a=1, b=3, c=-2. Applying the formula,

which yields

Solving it for both positive and negative values, we have x=4 and 1 which we obtained by factoring the equation. If there are rational roots, it's easy to acquire solution. But think the other case. Let's have an **quadratic formula example** with irrational roots.

**Quadratic Formula example without rational roots:**

*Next, let's solve the equation x^{2 }+ 3x – 2 = 0 using quadratic formula?

Here, if you try to factoring it, it's a cumbersome process. You need to go for quadratic formula method which is given below. Here we have a=1, b=3 and c=-2. Substituting all this values in the formula for quadratic equation,

which are not rational roots. We can approximate the results using calculator which would come around 0.56 and -3.56.

**Example for quadratic equation having complex roots:**

If the discriminant value i.e. b^{2}-4ac < 0, than it's square-root yields complex numbers and hence the roots of the quadratic equation are also complex. Let's see solution to one such equation.

*Solve the equation x^{2}+4x+5=0 using quadratic equation?

Here we have a=1, b=4, c=5. Clearly b^{2}-4ac = -4<0. We know that square-root of -1 is 'i'. Applying the formula, we have

which results in 2 complex roots, -2+2i and -2-2i. Thus the formula provides a effective solution to all kinds of quadratic equations.

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