Question

step-by-step how do i solve
1. (x + 5)(x - 2) = 0
2. x^2 - 3x + 27

Answer 1

\[\large \begin{array}{l}
(x + 5)(x - 2) = 0\\
{x^2} - 10 - 2x + 5x = 0\\
{x^2} + 3x - 10 = 0\\
\frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\\
\frac{{ - 3 \pm \sqrt {9 + 40} }}{2}\\
\frac{{ - 3 \pm 7}}{2}\\
{x_1} = - 5,{x_2} = 2
\end{array}\]

Answer 2

\[\large \begin{array}{l}
{x^2} - 3x + 27\\
\frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\\
\frac{{3 \pm \sqrt {9 - 108} }}{2}\\
It - is - negative - so - no - solutions - in - R
\end{array}\]

Answer 3

The root is negative, obviously :)