Question

In simplest radical form, what are the solutions to the quadratic eqaution 0=-3x^2-4x+5

Answer 1

Answer:

$x= -\frac{2+\sqrt{19} }{3} \textrm { and } x = -\frac{2-\sqrt{19} }{3}$

Step-by-step explanation:

We have to find the solution of the quadratic single variable equation as given below:

- 3x² - 4x + 5 = 0 ..... (1)

The left-hand side can not be factorized.

So, apply The Sridhar Acharya Formula, which gives if ax² + bx + c = 0,the the solutions of the quadratic equation are

$x=\frac{-b + \sqrt{b^{2} -4ac} }{2a}$ and, $x=\frac{-b - \sqrt{b^{2} -4ac} }{2a}$

So, from the equation (1), we can write  

$x= \frac{-(-4) +\sqrt{(-4)^{2}-4 \times (-3) \times 5 } }{2 \times (-3)}$ and

$x= \frac{-(-4) -\sqrt{(-4)^{2}-4 \times (-3) \times 5 } }{2 \times (-3)}$

⇒ $x=\frac{4+ 2\sqrt{19} }{-6} \textrm{ and } x=\frac{4-2\sqrt{19} }{-6}$

⇒ $x= -\frac{2+\sqrt{19} }{3} \textrm { and } x = -\frac{2-\sqrt{19} }{3}$ (Answer)