# How do you find the roots, real and imaginary, of y= 2x^2-15x-(4x+3)^2  using the quadratic formula?

Jun 30, 2016

x~~-1.69" and "x~~-0.38" to 2 decimal places")

#### Explanation:

You need to add like terms such that you and up with equation form:$\text{ } y = a {x}^{2} + b x + c$

To do this expand the brackets and simplify.
'~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Expanding the brackets}}$

Note that ${\left[- \left(4 x + 3\right)\right]}^{2} \text{ is not the same as } - \left[{\left(4 x + 3\right)}^{2}\right]$
The condition we have is$\text{ } - \left[{\left(4 x + 3\right)}^{2}\right]$

$\textcolor{b l u e}{\left(4 x + 3\right)} \textcolor{b r o w n}{\left(4 x + 3\right)}$

$\textcolor{b r o w n}{\textcolor{b l u e}{4 x} \left(4 x + 3\right) \textcolor{b l u e}{+ 3} \left(4 x + 3\right)}$

$16 {x}^{2} + 12 x \text{ "+12x+9" "=" } \textcolor{g r e e n}{16 {x}^{2} + 24 x + 9}$

,~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Substituting the expanded brackets and solving}}$

$\implies y = 2 {x}^{2} - 15 x - \left(\textcolor{g r e e n}{16 {x}^{2} + 24 x + 9}\right)$

Multiply everything inside the brackets by -1 and group like terms

$\implies y = 2 {x}^{2} - 16 {x}^{2} - 15 x - 24 x - 9$

$\implies y = - 14 {x}^{2} - 29 x - 9$
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From standard form we have:
$a = - 14$
$b = - 29$
$c = - 9$

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

becomes" "x=(+29+-sqrt((-29)^2-4(-14)(-9)))/(2(-14)

$x = \frac{29 \pm \sqrt{841 - 504}}{- 28}$

$x = \frac{29 \pm \sqrt{337}}{- 28} \text{ "larr 337" is a prime number}$

$\textcolor{b l u e}{x \approx - 1.69 \text{ and "x~~-0.38" to 2 decimal places}}$ 