How do you solve quadratic equations algebraically ?

Hint:We know that quadratic equations are the polynomial equations with degree 2 and one variable. The general form of this equation is \[a{x^2} + bx + c = 0\]. Here, $a$, $b$ and $c$ must be real numbers and $a$ cannot be zero. The values of $x$ satisfying the quadratic equation are the roots of the quadratic equation $\alpha $ and $\beta $.

Complete step by step answer:
We know that quadratic equations normally have two roots. The nature of roots can be real or imaginary. The solution of the quadratic equation can be given by the quadratic formula.Let us take a quadratic equation \[a{x^2} + bx + c = 0\]. The formula for its solution is given by:
$x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$, where, $D = {b^2} - 4ac$
Therefore, if $\alpha $ and $\beta $ are the roots of a quadratic equation, then they are given by
$\left( {\alpha ,\beta } \right) = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Here, the nature of the roots depends on the value of D. There are three types of roots depending on the value of D.The roots are real and unequal if the value of D is greater than zero.The roots are real and equal if the value of D is equal to zero.The roots are imaginary and unequal if the value of D is less than zero.Thus, we can find the roots and their nature by this method.

Note:We have seen how to find the roots and their nature by using quadratic formula. We can also find the sum and the product of both the roots. The sum of both roots is equal to the ratio of coefficient of $x$ to the coefficient of ${x^2}$with a negative sign which can be given as: $\alpha + \beta = - \dfrac{b}{a}$.And the product of both roots is equal to the ratio of the constant term to the coefficient of ${x^2}$ which can be given as: $\alpha \beta = \dfrac{c}{a}$.