As terms help compensate for a quadratic equation. It can indeed be solved in several ways, and the calculation of the equation can be defined as follows:
Quadratic Equation:
All quadratic equations can be stated as ax²+bx+c=0, where can not be equal to 0, but can.
Following are solutions to given equation:
[tex]\to[/tex] x² – 5x + 3 = 0
Comparing the value with the standard equation:
[tex]\to[/tex] ax² +bx +c = 0
[tex]\to[/tex] a= 1
[tex]\to[/tex] b=-5
[tex]\to[/tex] c=3
Appling the value into the Shreedhara Acharya's formula:
[tex]\to x = \frac {- b \pm \sqrt{ b^2 - 4ac } } { 2a}\\\\\to x = \frac {- (-5) \pm \sqrt{ (-5)^2 - 4\times 1 \times 3 } } { 2\times 1}\\\\\to x = \frac {+5 \pm \sqrt{ 25 - 12} } { 2}\\\\\to x = \frac {5 \pm \sqrt{ 13} } { 2}\\\\\to x = \frac {5 \pm 3.6} { 2}\\\\\to x = \frac {5 + 3.6 } { 2} \ \ \ \ \ \ OR \ \ \ \ \ \ \ x = \frac {5 - 3.6} { 2}\\\\\to x = \frac {8.6 } { 2} \ \ \ \ \ \ OR \ \ \ \ \ \ \ x = \frac {1.4} { 2}\\\\\to x = 4.3 \ \ \ \ \ \ OR \ \ \ \ \ \ \ x =0.7\\\\[/tex]
Therefore, the value of x is "4.3, 0.7".
Find out more about the Quadratic Equation here:
brainly.com/question/2263981
