Solving a quadratic equation with the quadratic formula

Key concept: This equation does not factor nicely, so the quadratic formula is the best method. The main steps are identifying $a$, $b$, and $c$, substituting them carefully, and simplifying the radical exactly.

Step by step

Identify the coefficients

The equation is

$$x^2-8x-3=0.$$

Compare it to the standard form

$$ax^2+bx+c=0.$$

Here,

$$a=1,\quad b=-8,\quad c=-3.$$

Apply the quadratic formula

Use

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

Substitute the values:

$$x=\frac{-(-8)\pm\sqrt{(-8)^2-4(1)(-3)}}{2(1)}.$$

Now simplify step by step:

$$x=\frac{8\pm\sqrt{64+12}}{2}$$ $$x=\frac{8\pm\sqrt{76}}{2}.$$

Since $76=4\cdot 19$,

$$\sqrt{76}=2\sqrt{19}.$$

So

$$x=\frac{8\pm 2\sqrt{19}}{2}=4\pm\sqrt{19}.$$

Write the final answers

The solutions are

$$4-\sqrt{19},\;4+\sqrt{19}.$$

If your system asks for a comma-separated list, enter:

$$4-\sqrt{19},\;4+\sqrt{19}$$

Why this method works

The quadratic formula works for every quadratic equation, whether or not it factors. That makes it especially useful when the constant term is small but no obvious integer factors appear. The discriminant, $b^2-4ac$, also tells you whether the solutions are real or complex. In this problem, the discriminant is positive, so there are two real solutions.

Common accuracy check

A quick check is to plug each root back into the original equation or to verify that the sum of the roots equals $8$ and the product equals $-3$, consistent with Vieta's formulas for $x^2-8x-3=0$.

Pitfall alert

The biggest mistake is assigning the coefficients incorrectly after comparing with standard form. Here, the coefficient of $x$ is $-8$, not $8$, and the constant term is $-3$, not $3$. Another frequent error is forgetting that $-b$ becomes positive when $b=-8$. Students also sometimes simplify $\sqrt{76}$ incorrectly by stopping at a decimal approximation too early. Keep the radical exact until the final answer unless the instructions ask for decimals.

Try different conditions

If the equation were $x^2-8x+3=0$, the discriminant would change to $64-12=52$, and the roots would be $x=\frac{8\pm\sqrt{52}}{2}=4\pm\sqrt{13}$. That variant shows how a sign change in the constant term affects the discriminant and the simplified radical. If the equation were instead $2x^2-8x-3=0$, then $a$ would be 2, and the denominator in the quadratic formula would change to 4, which alters the final answers again.

Further reading

quadratic formula, discriminant, standard form