Solve -6+5x-(4)+x^2=2

$$-6+5x-(4)+ { x }^{ 2 } =2$$

Solve for x

$$x = \frac{\sqrt{73} - 5}{2} \approx 1.772001873$$

$$x=\frac{-\sqrt{73}-5}{2}\approx -6.772001873$$

Steps Using the Quadratic Formula

Subtract 4 from -6 to get -10.

$$-10+5x+x^{2}=2$$

Subtract 2 from both sides.

$$-10+5x+x^{2}-2=0$$

Subtract 2 from -10 to get -12.

$$-12+5x+x^{2}=0$$

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.

$$x^{2}+5x-12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 5 for b, and -12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-5±\sqrt{5^{2}-4\left(-12\right)}}{2}$$

Square 5.

$$x=\frac{-5±\sqrt{25-4\left(-12\right)}}{2}$$

Multiply -4 times -12.

$$x=\frac{-5±\sqrt{25+48}}{2}$$

Add 25 to 48.

$$x=\frac{-5±\sqrt{73}}{2}$$

Now solve the equation x=\frac{-5±\sqrt{73}}{2} when ± is plus. Add -5 to \sqrt{73}.

$$x=\frac{\sqrt{73}-5}{2}$$

Now solve the equation x=\frac{-5±\sqrt{73}}{2} when ± is minus. Subtract \sqrt{73} from -5.

$$x=\frac{-\sqrt{73}-5}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{73}-5}{2} x=\frac{-\sqrt{73}-5}{2}$$

Steps for Completing the Square

Subtract 4 from -6 to get -10.

$$-10+5x+x^{2}=2$$

Add 10 to both sides.

$$5x+x^{2}=2+10$$

Add 2 and 10 to get 12.

$$5x+x^{2}=12$$

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.

$$x^{2}+5x=12$$

Divide 5, the coefficient of the x term, by 2 to get \frac{5}{2}. Then add the square of \frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+5x+\left(\frac{5}{2}\right)^{2}=12+\left(\frac{5}{2}\right)^{2}$$

Square \frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+5x+\frac{25}{4}=12+\frac{25}{4}$$

Add 12 to \frac{25}{4}.

$$x^{2}+5x+\frac{25}{4}=\frac{73}{4}$$

Factor x^{2}+5x+\frac{25}{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}.

$$\left(x+\frac{5}{2}\right)^{2}=\frac{73}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{2}\right)^{2}}=\sqrt{\frac{73}{4}}$$

Simplify.

$$x+\frac{5}{2}=\frac{\sqrt{73}}{2} x+\frac{5}{2}=-\frac{\sqrt{73}}{2}$$

Subtract \frac{5}{2} from both sides of the equation.

$$x=\frac{\sqrt{73}-5}{2} x=\frac{-\sqrt{73}-5}{2}$$