FIRERED21GO
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Please help solve this:

27x^2 + 15x + 10 = 0

I don't get what I need to do.​

Answer :

DᴀƦᴋPᴀƦᴀᴅᴏXGO

[tex]▪▪▪▪▪▪▪▪▪▪▪▪▪  {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪[/tex]

Let's solve for x :

  • [tex]27 {x}^{2} + 15x + 10 = 0[/tex]

Using quadratic formula ~

[tex] \boxed{ \frac{ - b \pm \sqrt{b {}^{2 } - 4ac}}{2a} }[/tex]

where,

  • b = 15

  • a = 27

  • c = 10

let's find the roots ~

  • [tex] \dfrac{ - 15 \pm \sqrt{( {15})^{2} - (4 \times 27 \times 10)} }{(2 \times 27)} [/tex]

  • [tex] \dfrac{ - 15 \pm \sqrt{225 - 1080} }{54}[/tex]

  • [tex] \dfrac{ - 15 \pm \sqrt{ - 855} }{54} [/tex]

  • [tex] \dfrac{ - 15 \pm3 \sqrt{95} \: i}{54}[/tex]

  • [tex] \dfrac{ - 15}{54} \pm \dfrac{3 \sqrt{95} }{54} i[/tex]

  • So, the required roots are ~

[tex]x = - \dfrac{5}{18} + \dfrac{3 \sqrt{95} }{54} [/tex]

and

[tex]x = - \dfrac{5}{18} - \dfrac{3 \sqrt{95} }{54} [/tex]

There is no real roots, both the roots are imaginary ~

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