Solve the quadratic equation ${x^2} - 3(x + 3) = 0$.

Hint: Solving a quadratic equation means finding its roots. There are three methods to find the roots of a quadratic equation. The methods are the Factorization method, the Formula method, and the Completing the square method. Here we will use the formula method.

Formula used:
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

Complete step-by-step answer:
The formula method states that, if the quadratic equation is of the form $a{x^2} + bx + c = 0$ ,then the roots of the quadratic equation are given by
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
The given quadratic equation is ${x^2} - 3(x + 3) = 0$.
Simplifying this quadratic equation,
${x^2} - 3(x + 3) = 0$
${x^2} - 3x - 9 = 0$
Comparing this equation with the standard quadratic equation $a{x^2} + bx + c = 0$ ,
$a = 1,b = - 3,c = - 9$
Substituting these values in the formula,
$x = \dfrac{{ - ( - 3) \pm \sqrt {{{( - 3)}^2} - 4(1)( - 9)} }}{{2(1)}}$
Simplifying and solving it further,
$x = \dfrac{{3 \pm \sqrt {9 + 4(9)} }}{2}$
$x = \dfrac{{3 \pm \sqrt {9 + 36} }}{2}$
$x = \dfrac{{3 \pm \sqrt {45} }}{2}$
We can write $45 = 3 \times 3 \times 5$ ,
$x = \dfrac{{3 \pm \sqrt {3 \times 3 \times 5} }}{2}$
Taking $3$ common from the square root,
$x = \dfrac{{3 \pm 3\sqrt 5 }}{2}$
The value of $\sqrt 5 = 2.2360$.
Substituting this value in the equation,
$x = \dfrac{{3 \pm (3 \times 2.2360)}}{2}$
Multiplying and adding,
$x = \dfrac{{3 \pm 6.708}}{2}$
We can say that,
$x = \dfrac{{3 + 6.708}}{2}$ or $x = \dfrac{{3 - 6.708}}{2}$
$x = \dfrac{{9.708}}{2}$ or $x = \dfrac{{ - 3.708}}{2}$
$x = 4.854$ or $x = - 1.854$
The roots of the quadratic equation ${x^2} - 3(x + 3) = 0$ are $4.854$ and $ - 1.854$ .
But we are asked to find the roots up to two significant figures only.
A significant value is the number of digits after a decimal point.
The value of $x$ up to two significant figures is $4.85$ and $ - 1.85$ .
Therefore, the correct answer of ${x^2} - 3(x + 3) = 0$ is $4.85$ and $ - 1.85$ .

Note: The $\sqrt {{b^2} - 4ac} $ term in the formula to find the roots of a quadratic equation is known as determinant. If $\sqrt {{b^2} - 4ac} = 0$, then the roots of the quadratic equation are real and equal. If $\sqrt {{b^2} - 4ac} > 0$ ,then the roots of the quadratic equation are real and unequal. If $\sqrt {{b^2} - 4ac} < 0$ , then the roots of the quadratic equation are not real i.e., the quadratic equation has imaginary roots.