Question

2x²-4x+5 find the zeros​

Answer 1

This equation is similar to equation in $ax^{2}+bx+c$

W. R. T to standard form

$a=2 \\b=-4\\c=5$

Using formula,

$\large\green{\boxed{\orange{\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}}}}$

On putting values,

=$\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$

=$\dfrac{-(-4)\pm\sqrt{(-4)^{2}-4\times2\times5}}{2\times2}$

=$\dfrac{4\pm\sqrt{(16-40}}{4}$

=$\dfrac{4\pm\sqrt{-24}}{4}$

=${\underline{\boxed{\red{\frac{4+\sqrt{-24}}{4}, {\frac{4-\sqrt{-24}}{4}}}}}}$

$<marquee scrollamount="1300">♥Answer by Rishabh♥</marquee>$

Answer 2

Answer:

p(x)=2x²-4x+5

a=2

b=-4

c=5

using quadratic formula

D=b²-4ac

D=(-4)²-4(2)(5)

=16-40

=-24

since, D<0

therefore, no real roots exist

x=(-b+√D)/2a. &. (-b-√D)/2a

x=(4+√-24)/4. & (4-√24)/4