The equation $2^x - x^2 = 0$ has

We need to find how many values of $x$ satisfy: $2^x = x^2$

In other words, where does the exponential function $2^x$ (red line) meet the parabola $x^2$ (blue line)?

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Let's plug in simple integer values and compare $2^x$ with $x^2$:

$x$	$2^x$	$x^2$	$2^x - x^2$
$-1$	$0.5$	$1$	$-0.5$
$0$	$1$	$0$	$+1$
$1$	$2$	$1$	$+1$
$2$	$4$	$4$	$0$ ✅
$3$	$8$	$9$	$-1$
$4$	$16$	$16$	$0$ ✅
$5$	$32$	$25$	$+7$

So we already found two solutions: $x = 2$ and $x = 4$.

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Now let's check for negative values of $x$.

For very negative $x$ (say $x = -10$), $2^x$ becomes tiny (close to $0$), but $x^2$ becomes huge ($100$). So $x^2 > 2^x$.

At $x = 0$: $2^x = 1 > 0 = x^2$, so $2^x > x^2$.

Since $x^2$ is above $2^x$ for very negative $x$, but $2^x$ is above $x^2$ at $x = 0$, the two curves must cross somewhere in between. This is guaranteed by the Intermediate Value Theorem — if one function overtakes the other, they must have been equal at some point.

This gives us a third solution in the negative region (approximately $x \approx -0.77$).

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For $x > 4$: $2^x$ grows exponentially while $x^2$ grows only as a polynomial, so $2^x$ stays above $x^2$ forever — no more crossings.

For $x < -0.77$ going further left: $x^2$ keeps growing while $2^x$ keeps shrinking toward $0$ — no more crossings.

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The equation $2^x - x^2 = 0$ has $3$ real solutions:

$x = 2, \quad x = 4, \quad \text{and} \quad x \approx -0.77$