Solve $x^2+19x+90=0$ using the quadratic formula

Key takeaway: This is a standard quadratic equation problem. The key is to identify $a$, $b$, and $c$, then substitute them carefully into the quadratic formula without skipping signs.

We start with

$x^2+19x+90=0$

For a quadratic in the form $ax^2+bx+c=0$, the quadratic formula is

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

Here,

• $a=1$
• $b=19$
• $c=90$

Substitute these values:

$x=\frac{-19\pm\sqrt{19^2-4(1)(90)}}{2(1)}$

$x=\frac{-19\pm\sqrt{361-360}}{2}$

$x=\frac{-19\pm\sqrt{1}}{2}$

$x=\frac{-19\pm 1}{2}$

Now split into the two solutions:

$x=\frac{-19+1}{2}=\frac{-18}{2}=-9$

$x=\frac{-19-1}{2}=\frac{-20}{2}=-10$

So the exact values of $x$ are:

$\boxed{x=-9 \text{ and } x=-10}$

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Pitfalls the pros know 👇 The most common mistake is sign handling: students often write $b=-19$ because they see the term $+19x$. For $ax^2+bx+c=0$, the coefficient $b$ is the number multiplying $x$, including its sign, so here $b=19$. Also, don’t forget to simplify the discriminant before taking the square root; $361-360=1$ makes the roots especially neat.

What if the problem changes? If the constant term changed, the same method would still work. For example, if the equation were $x^2+19x+89=0$, the discriminant would become $361-356=5$, so the exact answers would be irrational. If the leading coefficient were not $1$, you would still use the same formula, just with the new $a$ value plugged in carefully.

Tags: quadratic formula, discriminant, factorization