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?
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)
The quadratic formula statest that : $\displaystyle x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ where a is the coefficient of $\displaystyle x^2$, b is the coefficient of $\displaystyle x$ and c is the last term.
Plugging into the quadratic formula : $\displaystyle x=\frac{-3\pm \sqrt{(-3)^2-4(1)(-2)}}{2}\implies x=\frac{-3\pm\sqrt{17}}{2}$.
Derivation of the quadratic formula:
$\displaystyle ax^2+bx+c=0$
Subtract both sides by c.
$\displaystyle ax^2+bx=-c$
Divide through by a.
$\displaystyle x^2+\frac{b}{a}x=-\frac{c}{a}$
Complete the square on the left side (to keep equation balanced, add the new term to the right side as well)
$\displaystyle x^2+\frac{b}{a}x+\left(\frac{b}{2a}\right)^2=\frac {b^2}{4a^2}-\frac{c}{a}$
The left side becomes a perfect square. Combine the terms on the right side of the equation:
$\displaystyle \left(x+\frac{b}{2a}\right)^2=\frac{b^2}{4a^2}-\frac{c}{a}\cdot\frac{(4a)}{(4a)}\implies \left(x+\frac{b}{2a}\right)^2=\frac{b^2-4ac}{4a^2}$
Take the square root of both sides:
$\displaystyle x+\frac{b}{2a}=\pm\sqrt{\frac{b^2-4ac}{4a^2}}$
$\displaystyle \implies x+\frac{b}{2a}=\pm\frac{\sqrt{b^2-4ac}}{2a}$
Solve for x:
$\displaystyle x=-\frac{b}{2a}\pm\frac{\sqrt{b^2-4ac}}{2a}$
$\displaystyle \color{red}\boxed{\therefore x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}}$