Verified Answer :

iloveonedirection

\(2x^2 + 6x + 5 = 0\)

\(x= \frac{-b+/- \sqrt{b^2 -4ac} }{2a}\)

\(x= \frac{-(6)+/- \sqrt{(6)^2 -4(2)(5)} }{2(2)}\)

\(x= \frac{-6+/- \sqrt{36 -40} }{4}\)

\(x= \frac{-6+/- \sqrt{-4} }{4}\)

\(x= \frac{-6+/- i\sqrt{4} }{4}\)

\(x= \frac{-6+/- 2i }{4}\)

NO REAL ROOTS

\(2x^2+6x+5=0\ \ \ \ /:2\\\\x^2+3x+2.5=0\\\\x^2+2x\cdot1.5=-2.5\\\\x^2+2x\cdot1.5+1.5^2-1.5^2=-2.5\\\\(x+1.5)^2-2.25=-2.5\\\\(x+1.5)^2=-2.5+2.25\\\\(x+1.5)^2=-0.25 < 0\ aren't\ real\ roots\ because\ for\ x\in\mathbb{R},\ a^2\geq0\)

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