# How do you solve 4x^2 +4x = 15 using the quadratic formula?

Apr 11, 2018

$x = \frac{3}{2} , \frac{5}{2}$.

#### Explanation:

First Of All, Convert the Equation to It's General Form $a {x}^{2} + b x + c = 0$.

So We have,

$\textcolor{w h i t e}{\times x} 4 {x}^{2} + 4 x = 15$

$\Rightarrow 4 {x}^{2} + 4 x - 15 = 0$ [Subtract $15$ from both sides.]

So, Comparing the Equation with the General Form, We get,

$a = 4 , b = 4 , c = - 15$.

So, Let's Find the Discriminat.

$D = {b}^{2} - 4 a c = {4}^{2} - 4 \cdot 4 \cdot \left(- 15\right) = 16 + 240 = 256$

As $D > 0$, we will get two roots which are real and distinct.

Now Use The Quadratic Formula or Sridhar Acharya's Rule (whatever you may call it in your country).

$\alpha = \frac{- b + \sqrt{D}}{2 a} = \frac{- 4 + \sqrt{256}}{2 \cdot 4} = \frac{- 4 + 16}{8} = \frac{3}{2}$

And $\beta = \frac{- b - \sqrt{D}}{2 a} = \frac{- 4 - \sqrt{256}}{2 \cdot 4} = \frac{- 4 - 16}{8} = \frac{5}{2}$

So, $x = \frac{3}{2} , \frac{5}{2}$

Hope this helps.

Apr 11, 2018

$\frac{3}{2} \text{ or} - \frac{5}{2}$

#### Explanation:

Make the expression equal to zero:

$4 {x}^{2} + 4 x - 15 = 0$

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

In our case we substitute:

$a = 4 , b = 4 , c = - 15$

So that the quadratic formula becomes:

$x = \frac{- 4 \pm \sqrt{{4}^{2} - 4 \cdot 4 \cdot \left(- 15\right)}}{2 \cdot 4}$

$= \frac{- 4 \pm \sqrt{16 + 240}}{8}$

$= \frac{- 4 \pm 16}{8}$

$= \frac{3}{2} \text{or} - \frac{5}{2}$

We can also solve this by factorising (here you need to guess)

$4 {x}^{2} + 4 x - 15 = 0$

$\left(2 x + 5\right) \left(2 x - 3\right) = 0$

Then make both parenthesis equal to zero to get the same answers:

$\left(2 x + 5\right) = 0 \text{ or } \left(2 x - 3\right) = 0$