Solve x^2-3x-3+21/4=0

$${ x }^{ 2 } -3x-3+ \frac{ 21 }{ 4 } = 0$$

Solve for x

$$x = \frac{3}{2} = 1\frac{1}{2} = 1.5$$

Steps Using the Quadratic Formula

Add -3 and \frac{21}{4} to get \frac{9}{4}.

$$x^{2}-3x+\frac{9}{4}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -3 for b, and \frac{9}{4} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times \frac{9}{4}}}{2}$$

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\times \frac{9}{4}}}{2}$$

Multiply -4 times \frac{9}{4}.

$$x=\frac{-\left(-3\right)±\sqrt{9-9}}{2}$$

Add 9 to -9.

$$x=\frac{-\left(-3\right)±\sqrt{0}}{2}$$

Take the square root of 0.

$$x=-\frac{-3}{2}$$

The opposite of -3 is 3.

$$x=\frac{3}{2}$$

Steps for Completing the Square

Add -3 and \frac{21}{4} to get \frac{9}{4}.

$$x^{2}-3x+\frac{9}{4}=0$$

Factor x^{2}-3x+\frac{9}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{3}{2}\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{2}\right)^{2}}=\sqrt{0}$$

Simplify.

$$x-\frac{3}{2}=0 x-\frac{3}{2}=0$$

Add \frac{3}{2} to both sides of the equation.

$$x=\frac{3}{2} x=\frac{3}{2}$$

The equation is now solved. Solutions are the same.

$$x=\frac{3}{2}$$