# 3. The Quadratic Formula

At the end of the last section (Completing the Square), we derived a general formula for solving quadratic equations. Here is that general formula:

For any quadratic equation `ax^2+ bx + c = 0`, the solutions for *x* can be found by using the quadratic formula:

`x=(-b+-sqrt(b^2-4ac))/(2a)`

The expression under the square root, `b^2− 4ac`, can tell us how many roots we'll get. (There's no magic here - just a consideration of what the square root of `b^2− 4ac` is.)

If ` b^2− 4ac = 0`, then we'll have **one root** only, `x = −b/(2a)`.

If ` b^2− 4ac > 0`, then we'll have **two roots**, one involving the "+" sign and the other involving the "−" sign in the formula.

If ` b^2− 4ac < 0`, then we'll have **no real roots**, since you cannot find the square root of a negative number.

The expression `b^2 − 4ac` is called the **discriminant** and in some books you will see it written with a Greek upper case Delta, like this `Delta = b^2 − 4ac`.

### Example 1

Solve `2x^2- 7x - 5 = 0` using the quadratic formula.

Answer

By inspection, we can see that: `a = 2`, `b = -7` and `c = -5`.

Substituting these into the quadratic formula, we get :

`x=(-b+-sqrt(b^2-4ac))/(2a)`

`=(-(-7)+-sqrt((-7)^2-4(2)(-5)))/(2(2))`

`=(7+-sqrt(49+40))/4`

`=(7+-sqrt(89))/4`

`=(7-sqrt89)/4 or (7+sqrt89)/4`

`=-0.6085 or 4.108`

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### Example 2

Solve `2x^2= 4x + 3`

Answer

Firstly, in order to identify *a*, *b* and *c* , we must re-arrange the expression in the proper form ie. all the terms to the left, leaving zero on the right:

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

Only then, we can see that:

`a = 2`, `b = -4` and `c = -3`

Substituting these into the quadratic formula, we get:

`x=(-b+-sqrt(b^2-4ac))/(2a)`

`x=(-(-4)+-sqrt((-4)^2-4(2)(-3)))/(2(2))`

`=(4+-sqrt(16+24))/4`

`=(4+-sqrt40)/4`

`=(4-sqrt40)/4 or (4+sqrt40)/4`

So

`x=-0.581 or 2.581`

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### Exercise

Solve `6r^2= 6r + 1` using the quadratic formula.

Answer

`6r^2=6r+1`

Rearranging so it's in quadratic form:

`6r^2-6r-1=0`

Using the Quadratic Equation for the variable `r`:

`r=(-b+-sqrt(b^2-4ac))/(2a)`

`=(-(-6)+-sqrt((-6)^2-4(6)(-1)))/(2(6))`

`=(6+-sqrt(36+24))/12`

`=(6+-sqrt60)/12`

`=(6-2sqrt15)/12 or (6+2sqrt15)/12`

`=(3-sqrt15)/6 or (3+sqrt15)/6`

So

`r=-0.145 or 1.145`

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