Question

What are the solutions of this quadratic equation? 6x2 + 6 = 12x + 18

Answer 1

Answer:

x=1+ sqrt 3

hope this helps

p.s. this is not my answer just clarifying it for every one.

Answer 2

The solution of the quadratic equation $6x^{2} + 6 = 12x + 18$ are $1 + \sqrt{3}$ and $1 - \sqrt{3}$.

Roots of Quadratic Equations:

• For the general quadratic equation, $ax^{2} + bx + c = 0$, the roots are given by

                        $x = \frac{ - b + \sqrt{D} }{2a}$  or $x = \frac{ - b - \sqrt{D} }{2a}$
       where, D = Discriminant = $b^{2} - 4 a c$.

How to solve the given quadratic equation?

1. Arrange the given equation in the general form.
   ∴ $6x^{2} + 6 - 12x -18 = 0$
   ∴ $6x^{2} - 12x -12 = 0$
2. Find the discriminant of the equation.
   ∴ comparing the coefficients from the general form,
   a = 6 , b = -12 & c = -12
   ∴ Discriminant of the given equation,
   ∴ $D = b^{2} - 4 a c = (-12)^2 - 4 (6) (-12) = 144 \ + \ 288 = 432$
3. Check the nature of roots.
   As D > 0, the roots of the equation are real and distinct.
4. Find the roots.
   ∴ $x = \frac{ - b + \sqrt{D} }{2a} = \frac{ - (-12) + \sqrt{432} }{2 \ (6)} = \frac{12 + 12 \sqrt{3} }{12} = 1 + \sqrt{3}$ &
   ∴ $x = \frac{ - b - \sqrt{D} }{2a} = \frac{ - (-12) - \sqrt{432} }{2 \ (6)} = \frac{12 - 12 \sqrt{3} }{12} = 1 - \sqrt{3}$

Thus the solutions of the given equations are $1 + \sqrt{3}$ and $1 - \sqrt{3}$.

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