# How do you find the roots, real and imaginary, of y= x^2 - 3x + 44-(3x-1)^2  using the quadratic formula?

Aug 9, 2017

You should know the 2 formulas:
1.) ${\left(a - b\right)}^{2} = {a}^{2} + {b}^{2} - 2 a b$
2.) Quadratic Formula: $\frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

#### Explanation:

Step 1.) Open the bracket term in question by formula no.1 and solve it further:

$y = {x}^{2} - 3 x + 44 - {\left(3 x - 1\right)}^{2}$
$y = {x}^{2} - 3 x + 44 - \left(9 {x}^{2} + 1 - 6 x\right)$

Now solvng it further:
$y = {x}^{2} - 3 x + 44 - 9 {x}^{2} - 1 + 6 x$
$y = - 8 {x}^{2} + 3 x + 43$

Step 2.) Now applying Quadratic formula:

$\frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
x= (-3+-sqrt(3^2-4(43)(-8)))/(2(-8)
$x = \frac{- 3 \pm \sqrt{9 + 1376}}{- 16}$
$x = \frac{- 3 \pm \sqrt{1385}}{- 16}$
$x = \frac{- 3 \pm 37.21}{- 16}$

Case 1:
when the sign is positive in between:
$x = \frac{- 3 + 37.21}{- 16}$
$x = \frac{34.21}{- 16}$
$x = - 2.13$

Case 2:
when the negative is positive in between:
$x = \frac{- 3 - 37.21}{- 16}$
$x = \frac{- 40.21}{- 16}$
$x = 2.51$

That's all, Thank You.