Solve Using the Quadratic Formula 24=x^2-4x+3

$24 = x^{2} - 4 x + 3$

Move all terms to the left side of the equation and simplify.

Move all the expressions to the left side of the equation.

Subtract $x^{2}$ from both sides of the equation.

$24 - x^{2} = - 4 x + 3$

Add $4 x$ to both sides of the equation.

$24 - x^{2} + 4 x = 3$

Subtract $3$ from both sides of the equation.

$24 - x^{2} + 4 x - 3 = 0$

Subtract $3$ from $24$ .

$- x^{2} + 4 x + 21 = 0$

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Substitute the values $a = - 1$ , $b = 4$ , and $c = 21$ into the quadratic formula and solve for $x$ .

$\frac{- 4 \pm \sqrt{4^{2} - 4 \cdot (- 1 \cdot 21)}}{2 \cdot - 1}$

Simplify.

Simplify the numerator.

Raise $4$ to the power of $2$ .

$x = \frac{- 4 \pm \sqrt{16 - 4 \cdot - 1 \cdot 21}}{2 \cdot - 1}$

Multiply $- 4$ by $- 1$ .

$x = \frac{- 4 \pm \sqrt{16 + 4 \cdot 21}}{2 \cdot - 1}$

Multiply $4$ by $21$ .

$x = \frac{- 4 \pm \sqrt{16 + 84}}{2 \cdot - 1}$

Add $16$ and $84$ .

$x = \frac{- 4 \pm \sqrt{100}}{2 \cdot - 1}$

Rewrite $100$ as $10^{2}$ .

$x = \frac{- 4 \pm \sqrt{10^{2}}}{2 \cdot - 1}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \frac{- 4 \pm 10}{2 \cdot - 1}$

Multiply $2$ by $- 1$ .

$x = \frac{- 4 \pm 10}{- 2}$

Simplify $\frac{- 4 \pm 10}{- 2}$ .

$x = 2 \pm 5$

The final answer is the combination of both solutions.

$x = 7, - 3$