Question

find the roots of the quadratic equation x^2-2x+3=0 by applying formula.​

Answer 1

Answer:

x²- 2x + 3 = 0

a = 1. , b = -2 , c = 3

Formula => -b +- √b²- 4ac / 2a

2 + √4 - 12 / 2. || 2 - √4 - 12 / 2

2 + √8 / 2 || 2 - √8 / 2

2 + 2 √2 / 2 || 2 - 2 √2 / 2

2 ( 1 + √2 ) / 2 || 2 ( 1 - √2 ) / 2

=> x = ( 1 + √2 ) , ( 1 - √2 )

All the best for exams.....

Answer 2

$\huge\bf\mathcal\pink{HeyDude}$
Answer:
$1 - \sqrt{2}$,
$1 + \sqrt{2}$

Step-by-Step Explanation :

Given :
${x}^{2} - 2x + 3 = 0$
Here,
$a \: = 1 \\ b = ( - 2) \\ c = 3$

$d \: \: = {b}^{2} \: - 4ac \\ = ( - 2)^{2} - 4(1)(3) \\ = 4 - 12 \\ = ( - 8)$

Now,

$x = \frac{ - b + \sqrt{d} }{2a} \\ x = \frac{-( - 2) + \sqrt{-8} }{2(1)} \\ x = \frac{ 2 + (-\sqrt[2]{2}{ } )}{2} \\ x = 1 - \sqrt{2}$

$x = \frac{ - b - \sqrt{d} }{2a} \\ x = \frac{-( - 2) - \sqrt{-8} }{2(1)} \\ x = \frac{ 2 - (-\sqrt[2]{2}{ }) }{2} \\ x = 1 + \sqrt{2}$
$\huge \bold \blue {Thanks..!}$ ​

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