Write the quadratic equation in standard form. Then solve using the quadratic formula.

2 - 3x + x^2 = 0

So in standard form, this is what I got:

x^2 + 3x - 2 = 0

How do I solve it using the quadratic formula?

2. Originally Posted by SarynJumail
Write the quadratic equation in standard form. Then solve using the quadratic formula.

2 - 3x + x^2 = 0

So in standard form, this is what I got:

x^2 + 3x - 2 = 0

How do I solve it using the quadratic formula?
-b+- SQRT((b^2)-4ac)
so..

-3 +- SQRT((3^2)-4(1)(-2))/(2)(1)

3. -3 +- SQRT(9) + (8) / (2)

-3 +- SQRT (17) / (2)

So where do I go from here?

4. Originally Posted by SarynJumail
-3 +- SQRT(9) + (8) / (2)

-3 +- SQRT (17) / (2)

So where do I go from here?
That should be it since sqrt(17) is not a perfect square so you get -3±√(17)/2, because you cannot cancel anything out, as your answer. Just by looking at the discriminate , in this case 17, you can determine that the roots of your equation will be real,unequal and irrational, also it has two x-intercepts if you are graphing it.( For future reference)

5. Originally Posted by SarynJumail
Write the quadratic equation in standard form. Then solve using the quadratic formula.

2 - 3x + x^2 = 0

So in standard form, this is what I got:

x^2 + 3x - 2 = 0

How do I solve it using the quadratic formula?
The quadratic formula statest that : $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ where a is the coefficient of $x^2$, b is the coefficient of $x$ and c is the last term.

Plugging into the quadratic formula : $x=\frac{-3\pm \sqrt{(-3)^2-4(1)(-2)}}{2}\implies x=\frac{-3\pm\sqrt{17}}{2}$.

6. Okay.
So -3 + SQRT of 7 / 2 and -3 - SQRT of 7/2.
That's right?

7. Originally Posted by SarynJumail
Okay.
So -3 + SQRT of 7 / 2 and -3 - SQRT of 7/2.
That's right?
$\frac{-3\pm \sqrt{17}}{2} = \frac{-3+\sqrt{17}}{2} \ or \ \frac{-3-\sqrt{17}}{2}$.

I hope that this is clear.

8. Derivation

$ax^2+bx+c=0$

Subtract both sides by c.

$ax^2+bx=-c$

Divide through by a.

$x^2+\frac{b}{a}x=-\frac{c}{a}$

Complete the square on the left side (to keep equation balanced, add the new term to the right side as well)

$x^2+\frac{b}{a}x+\left(\frac{b}{2a}\right)^2=\frac {b^2}{4a^2}-\frac{c}{a}$

The left side becomes a perfect square. Combine the terms on the right side of the equation:

$\left(x+\frac{b}{2a}\right)^2=\frac{b^2}{4a^2}-\frac{c}{a}\cdot\frac{(4a)}{(4a)}\implies \left(x+\frac{b}{2a}\right)^2=\frac{b^2-4ac}{4a^2}$

Take the square root of both sides:

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

Solve for x:

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

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