Question

Solve the following quadratic equation using the quadratic formula. Separate multiple answers with a comma if necessary.

Answer 1

Answer:

$y^2 -4y +6=0$

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

Where $a = 1, b= -4 ,c =6$

And replacing we got:

$y = \frac{-(-4) \pm \sqrt{4^2 -4(1)(6)}}{2*1}$

And solving we got:

$y = \frac{4 \pm \sqrt{-8}}{2} =2 \pm 2\sqrt{2} i$

Where $i =\sqrt{-1}$

And the possible solutions are:

$y_1=2 + 2\sqrt{2} i , y_2 = 2 - 2\sqrt{2} i$

Step-by-step explanation:

For this case we use the equation given by the image and we have:

$-y^2 +4y -6=0$

We can rewrite the last expression like this if we multiply both sides of the equation by -1.

$y^2 -4y +6=0$

Now we can use the quadratic formula given by:

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

Where $a = 1, b= -4 ,c =6$

And replacing we got:

$y = \frac{-(-4) \pm \sqrt{4^2 -4(1)(6)}}{2*1}$

And solving we got:

$y = \frac{4 \pm \sqrt{-8}}{2} =2 \pm 2\sqrt{2} i$

Where $i =\sqrt{-1}$

And the possible solutions are:

$y_1=2 + 2\sqrt{2} i , y_2 = 2 - 2\sqrt{2} i$