$\displaystyle -3x^2+6x-12=0$

This function cannot be factor therefore, use the quadratic equation.

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

Since the original equation is in the form $\displaystyle ax^2+bx+c$ where 

$\displaystyle a=-3, b=6, c=-12$.

Therefore,

$\displaystyle \frac{-6\pm\sqrt{6^2-4(-3)(-12)}}{2(-3)}$

$\displaystyle \frac{-6\pm\sqrt{36-144}}{-6}$

$\displaystyle \frac{-6\pm\sqrt{-108}}{-6}$

Since the value of the discriminant (the value beneath the square root symbol) is negative, this function has no real solutions.

$\displaystyle x^2+8x-2=0$

This particular function cannot be factored therefore, use the quadratic formula to solve.

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

Since the function is in the form $\displaystyle ax^2+bx+c$ where 

$\displaystyle a=1,b=8, c=-2$

the quadratic formula becomes as follows.

$\displaystyle \frac{-8\pm {\sqrt{8^2-4(1)(-2)}}}{2(1)}$

$\displaystyle \frac{-8\pm {\sqrt{64+8}}}{2}$

$\displaystyle \frac{-8\pm {\sqrt{72}}}{2}$

$\displaystyle \frac{-8\pm {\sqrt{9}\times\sqrt{4}\times\sqrt{2}}}{2}$

$\displaystyle \frac{-8\pm {3\times2\times\sqrt{2}}}{2}$

$\displaystyle {-4\pm {3\sqrt{2}}}$

To solve this problem, you must first rewrite the equation into  $\displaystyle ax^2+bx+c$ form (quadratic form). 

After this you plug the numbers into the following quadratic equation:

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

Which upon doing  you get:

$\displaystyle x=\frac{-4\pm \sqrt{16-8}}{2}$ 

This simplifies to:

$\displaystyle x=-2\pm\sqrt{2}$

$\displaystyle 2x^2 +3x - 7 = 0$

Recall the general form of a quadratic, 

$\displaystyle ax^2 +bx+c = 0$

The solution set has the form, 

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

For our particular case, $\displaystyle a = 2$, $\displaystyle b = 3$, and $\displaystyle c = -7$

$\displaystyle x = \frac{-3 \pm \sqrt{3^2 - 4(2)(-7)}}{2(2)}$

$\displaystyle x = \frac{-3 \pm \sqrt{65}}{4}$

$\displaystyle x = \left\{-\frac{3}{4} + \frac{\sqrt{65}}{4}, -\frac{3}{4} - \frac{\sqrt{65}}{4} \right\}$

$\displaystyle 2x^2-12x+16=0$

First, simplify the equation, so that the numbers are easier to work with. We can see that we can factor a 2 from each term.

$\displaystyle 2(x^2-6x+8)=0$

Now we can divide both sides by 2, to further simplify.

$\displaystyle \frac{2}{2}(x^2-6x+8)=\frac{0}{2}$

$\displaystyle x^2-6x+8 = 0$

Now that we have simplified we can apply the quadratic formula. 

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

where a, b, and c are the constants defined as follows:

$\displaystyle ax^2+bx+c = 0$

This means our a is 1, our b is -6, and our c is 8.

Finally, lets plug the numbers into the formula:

$\displaystyle \frac{-(-6)\pm \sqrt{(-6)^2-4(1)(8)}}{2(1)}$

$\displaystyle \frac{6\pm \sqrt{36-32}}{2}$

$\displaystyle \frac{-6\pm \sqrt{4}}{2}$

$\displaystyle \frac{-6\pm 2}{2}$

$\displaystyle \frac{-4}{2}$ and $\displaystyle \frac{-8}{2}$

or more simply:

$\displaystyle 2, 4$ 

These are the roots of the equation.

$\displaystyle 4x^2-9x+3$

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

where a, b, and c are the constants defined as follows:

$\displaystyle ax^2+bx+c = 0$

This means our a is 1, our b is -6, and our c is 8.

Finally, lets plug the numbers into the formula:

$\displaystyle \frac{-(-9)\pm \sqrt{(-9)^2-4(4)(3)}}{2(4)}$

$\displaystyle \frac{9\pm \sqrt{81-48}}{8}$

$\displaystyle \frac{9\pm \sqrt{33}}{8}$

These are the roots of the equation.

$\displaystyle x =\frac{9\pm \sqrt{33}}{8}$

Remember that using $\displaystyle \pm$ is the exact same as writing:

$\displaystyle x =\frac{9+ \sqrt{33}}{8}, x =\frac{9- \sqrt{33}}{8}$

The quadratic equation is:

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

$\displaystyle \\a=3 \\b=-9 \\c=2$

Therefore:

$\displaystyle \frac{-(-9)\pm \sqrt{(-9)^2-4(3)(2)}}{2(3)}$

Which gives the answer:

$\displaystyle x=\frac{9\pm \sqrt{57}}{6}$

For a quadratic function 

$\displaystyle ax^2+bx+c$

the quadratic formula states that

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

Using the formula for our function, we get

$\displaystyle x=\frac{-12\pm \sqrt{(12)^2-4(3)(15)}}{2(3)}=\frac{-12\pm \sqrt{-36}}{6}$

Notice that we have a negative under the square root. This means that we must use the imaginary number 

$\displaystyle i=\sqrt{-1}$

and our roots will be imaginary.

Simplifying using i, we get

$\displaystyle x=\frac{-12\pm 6i}{6}=-2\pm i$

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

For this problem

a=1, the coefficient on the x^2 term

b=7, the coefficient on the x term

c=7, the constant term

$\displaystyle \frac{-7\pm \sqrt{7^2-4(1)(7)}}{2(1)}\Rightarrow \frac{-7\pm \sqrt{21}}{2}$

Write the quadratic equation that applies for $\displaystyle y=ax^2+bx+c$.

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

There is no $\displaystyle b$ term.   Substitute the known coefficients from the polynomial.

$\displaystyle x=\frac{\pm \sqrt{-4(15)(1)}}{2(15)}$

Simplify the numerator and denominator.

$\displaystyle x=\frac{ \pm\sqrt{-60}}{30} = \pm\frac{\sqrt{-1} \cdot 2 \sqrt{15}}{30} = \pm \frac{i\sqrt{15}}{15}$

The roots will be imaginary.

One of the possible answers is:  $\displaystyle -\frac{i\sqrt{15}}{15}$