# How do you find the zeros, real and imaginary, of y=x^2-3x+29 using the quadratic formula?

Jan 1, 2016

#### Answer:

The roots are complex : ${X}_{1} = 1.5 + 5.172 i \mathmr{and} {X}_{2} = 1.5 - 5.172 i$ where i is an imaginary number ($i = \sqrt{-} 1$)

#### Explanation:

Comparing the above equation with General Quadratic equation $a {x}^{2} + b x + c$ we get a =1 ; b=-3; c=29 Now we see here ${b}^{2} - 4 \cdot a \cdot c = - 107$ If ${b}^{2} - 4 \cdot a \cdot c < 0$ then the roots are complex number.
Roots are ($- \frac{b}{2 \cdot a} + \sqrt{{b}^{2} - 4 \cdot a \cdot c}$/$\left(2 \cdot a\right)$) and ($- \frac{b}{2 \cdot a} - \sqrt{{b}^{2} - 4 \cdot a \cdot c}$/$\left(2 \cdot a\right)$) or $\frac{3}{2}$+$\sqrt{9 - 116}$ / $2 = 1.5 + 5.172 i$ and $\frac{3}{2}$ - $\sqrt{9 - 116}$ / #2 = 1.5 - 5.172i [Answer]