Question

find the nature of root of quadratic equation
${y}^{2} + 6y - 2 = 0$
using
$Δ = {b}^{2} - 4ac$
​

Answer 1

Answer:

check the attachment for the answer ...

hope this helps you...

Answer 2

Answer:

$\huge\underline{ \underline{ \red{your \: \: answer}}}$

Step-by-step explanation:

$\sf\pink{y = \frac{ - b±\sqrt{ {b}^{2} - 4ac} }{2a} } \\ \sf\blue{ = > y = \frac{ - 6± \sqrt{36 - 4 \times 1 \times - 2} }{2 \times 1} } \\ \sf\red{ = > y = \frac{ - 6± \sqrt{36 + 8} }{2} } \\ \sf\orange{ = > y = \frac{ - 6± \sqrt{44} }{2} } \\ \sf\green{ = > y = \frac{ - 6±2 \sqrt{11} }{2} } \\ \sf\pink{y = \frac{ - 6 + 2 \sqrt{11} }{2} = - 3 + \sqrt{11} } \\ \sf\purple{y = \frac{ - 6 - 2 \sqrt{11} }{2} = - 3 - \sqrt{11} } \\ \\ \large\boxed{ \fcolorbox{pink}{yellow}{hope \: it \: help \: you}}$