Question

Find roots of equation root3x^2 +11x+root28

Answer 1

$\sqrt{3} x^2 + 11x + \sqrt{28} = 0$
use formula
$x = \frac{b+ \sqrt{b^2-4ac} }{2a}, \frac{b - \sqrt{b^2-4ac} }{2a}$
for equation 
$ax^2+bx+c = 0$
Now
[tex]a = \sqrt{3}, b = 11,c = \sqrt{28} \\ x = \frac{11+ \sqrt{11^2-4(\sqrt{3})(\sqrt{28})} }{2\sqrt{3}}, \frac{11- \sqrt{11^2-4(\sqrt{3})(\sqrt{28})} }{2\sqrt{3}} \\ x = -0.5243355975235029,-5.826517363558263[/tex]

Answer 2

it's your answer..........