What are the solutions of the quadratic equation

Ответ:

nani122

04.01.2022

A quadratic equation ax^2+bx+c=0 ....[1]

then the solution is given by:

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

Given the equation:

3x^2-x =11

Subtract 11 from both sides we have;

3x^2-x-11 =0

On comparing with [1] we have;

a = 3 , b =-1 and c =-11

Substitute these we have;

$x = \frac{-(-1) \pm \sqrt{(-1)^2-4(3)(-11)}}{2(3)}$

⇒ $x = \frac{1 \pm \sqrt{1+132}}{6}$

⇒ $x = \frac{1 \pm \sqrt{133}}{6}$

Therefore, the solution for the given equations are:

$x = \frac{1 + \sqrt{133}}{6}$ , $\frac{1 -\sqrt{133}}{6}$

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