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

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

Solve for x

$$x = \frac{3 \sqrt{2} + 3}{2} \approx 3.621320344$$

$$x=\frac{3-3\sqrt{2}}{2}\approx -0.621320344$$

Steps Using the Quadratic Formula

Add x^{2} to both sides.

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

Subtract 2x from both sides.

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

Subtract 3 from both sides.

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

Subtract 3 from \frac{3}{4} to get -\frac{9}{4}.

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

Combine -x and -2x to get -3x.

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

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

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

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}.

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

Square -3.

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

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

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

Add 9 to 9.

$$x=\frac{-\left(-3\right)±\sqrt{18}}{2}$$

Take the square root of 18.

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

The opposite of -3 is 3.

$$x=\frac{3±3\sqrt{2}}{2}$$

Now solve the equation x=\frac{3±3\sqrt{2}}{2} when ± is plus. Add 3 to 3\sqrt{2}.

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

Now solve the equation x=\frac{3±3\sqrt{2}}{2} when ± is minus. Subtract 3\sqrt{2} from 3.

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

The equation is now solved.

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

Steps for Completing the Square

Add x^{2} to both sides.

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

Subtract 2x from both sides.

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

Subtract \frac{3}{4} from both sides.

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

Subtract \frac{3}{4} from 3 to get \frac{9}{4}.

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

Combine -x and -2x to get -3x.

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

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

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

Divide -3, the coefficient of the x term, by 2 to get -\frac{3}{2}. Then add the square of -\frac{3}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.

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

Add \frac{9}{4} to \frac{9}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

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

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

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

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