# How do you find the zeros, real and imaginary, of  =5(3x-3)^2+13  using the quadratic formula?

Apr 19, 2017

$x = 1 \pm i \sqrt{\frac{13}{45}}$

#### Explanation:

$5 {\left(3 x - 3\right)}^{2} = 13 {i}^{2}$ implies
$45 {\left(x - 1\right)}^{2} = 13 {i}^{2}$
$x - 1 = \pm i \sqrt{\frac{13}{45}}$
$x = 1 \pm i \sqrt{\frac{13}{45}}$

Apr 19, 2017

Zeros are $1 + 0.5375 i$ and $1 - 0.5375 i$, two complex conjugate numbers.

#### Explanation:

According to quadratic formula, zeros of $a {x}^{2} + b x + c$ are given by $\frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

As function is $5 {\left(3 x - 3\right)}^{2} + 13$

=$5 \left(9 {x}^{2} - 18 x + 9\right) + 13$

= $45 {x}^{2} - 90 x + 45 + 13$

= $45 {x}^{2} - 90 x + 58$

And zeros are (-(-90)+-sqrt((-90)^2-4×45×58))/(2×45)

= $\frac{90 \pm \sqrt{8100 - 10440}}{90}$

= $1 \pm \frac{\sqrt{- 2340}}{90}$

= $1 \pm i \frac{48.37}{90}$

= 1+-i×0.5375

i.e. Zeros are $1 + 0.5375 i$ and $1 - 0.5375 i$, two complex conjugate numbers.