A quadratic equation can be defined as a equation of a second degree, which implies that it comprises of minimum one term that is squared. The definite form is ax² + bx + c = 0; where in x is an unknown variable and a,b,c are numerical coefficients.
Quadratics Equation Examples
Beneath are the illustrations of quadratic equations of the form (ax² + bx + c = 0)
 x² –x – 9 = 0
 5x² – 2x – 6 = 0
 3x² + 4x + 8 = 0
 x² +6x + 12 = 0
Examples of a quadratic equation with the absence of a ‘ C ‘ a constant term.
 x² – 9x = 0
 x² + 2x = 0
 6x² – 3x = 0
 5x² + x = 0
 12x² + 13x = 0
 11x² – 27x = 0
Following are the examples of a quadratic equation in factored form
 (x – 6)(x + 1) = 0 [ result obtained after solving is x² – 5x – 6 = 0]
 –3(x – 4)(2x + 3) = 0 [result obtained after solving is 6x² + 15x + 36 = 0]
 (x − 5)(x + 3) = 0 [result obtained after solving is x² − 2x − 15 = 0]
 (x – 5)(x + 2) = 0 [ result obtained after solving is x² – 3x – 10 = 0]
 (x – 4)(x + 2) = 0 [result obtained after solving is x² – 2x – 8 = 0]
 (2x+3)(3x – 2) = 0 [result obtained after solving is 6x² + 5x – 6]
Below are the examples of a quadratic equation with an absence of linear co – efficient ‘ bx’
 2x² – 64 = 0
 x² – 16 = 0
 9x² + 49 = 0
 2x² – 4 = 0
 4x² + 81 = 0
 x² – 9 = 0
Solving Quadratics by Factoring
 Begin with a equation of the form ax² + bx + c = 0
 Ensure that it is set to adequate zero.
 Factor the left hand side of the equation by assuming zero on the right hand side of the equation.
 Assign each factor equal to zero.
 Now solve the equation in order to determine the values of x.
Suppose if the main coefficient is not equal to one then deliberately, you have to follow a methodology in the arrangement of the factors.
2x²x6=0
(2x+3)(x2)=0
2x+3=0
x=3/2
x=2
Also, Sridhara wrote down rules for Solving Quadratic Equation, therefore the most common method of finding the roots of the quadratic equation is known as Sridharacharya rule.
For the given Quadratic equation of the form, \(ax^{2}+bx+ c = 0\)
Therefore the roots of the given equation can be found by:
\(x = \frac{b \pm \sqrt{b^{2}4ac}}{2a}\),
where \(\pm\) (one plus and one minus) represent two distinct roots of the given equation.
Lets Work Out: Example: \(3x^{2} – 5x + 2 = 0\) Solution: \(3x^{2} – 5x + 2 = 0\) Solving the quadratic equation using the above method: \(x= \frac{b \pm \sqrt{b^{2}4ac}}{2a}\) \(x = \frac{(5)\pm \sqrt{(5)^{2} 4 \times 3 \times 2}}{2 \times 3}\) \(x = \frac{5 \pm 1}{6}\) \(x = \frac{6}{6} \;\; or \;\; \frac{4}{6}\) or, \(x = 1 \;\; or \;\; \frac{2}{3}\)

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