Solve -x-5=0.5x^2+1.5x-2

$$-x-5=0.5 { x }^{ 2 } +1.5x-2$$

Solve for x

$$x=-3$$

$$x=-2$$

Steps Using the Quadratic Formula

Subtract 0.5x^{2} from both sides.

$$-x-5-0.5x^{2}=1.5x-2$$

Subtract 1.5x from both sides.

$$-x-5-0.5x^{2}-1.5x=-2$$

Add 2 to both sides.

$$-x-5-0.5x^{2}-1.5x+2=0$$

Add -5 and 2 to get -3.

$$-x-3-0.5x^{2}-1.5x=0$$

Combine -x and -1.5x to get -2.5x.

$$-2.5x-3-0.5x^{2}=0$$

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.

$$-0.5x^{2}-2.5x-3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -0.5 for a, -2.5 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-2.5\right)±\sqrt{\left(-2.5\right)^{2}-4\left(-0.5\right)\left(-3\right)}}{2\left(-0.5\right)}$$

Square -2.5 by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\left(-2.5\right)±\sqrt{6.25-4\left(-0.5\right)\left(-3\right)}}{2\left(-0.5\right)}$$

Multiply -4 times -0.5.

$$x=\frac{-\left(-2.5\right)±\sqrt{6.25+2\left(-3\right)}}{2\left(-0.5\right)}$$

Multiply 2 times -3.

$$x=\frac{-\left(-2.5\right)±\sqrt{6.25-6}}{2\left(-0.5\right)}$$

Add 6.25 to -6.

$$x=\frac{-\left(-2.5\right)±\sqrt{0.25}}{2\left(-0.5\right)}$$

Take the square root of 0.25.

$$x=\frac{-\left(-2.5\right)±\frac{1}{2}}{2\left(-0.5\right)}$$

The opposite of -2.5 is 2.5.

$$x=\frac{2.5±\frac{1}{2}}{2\left(-0.5\right)}$$

Multiply 2 times -0.5.

$$x=\frac{2.5±\frac{1}{2}}{-1}$$

Now solve the equation x=\frac{2.5±\frac{1}{2}}{-1} when ± is plus. Add 2.5 to \frac{1}{2} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{3}{-1}$$

Divide 3 by -1.

$$x=-3$$

Now solve the equation x=\frac{2.5±\frac{1}{2}}{-1} when ± is minus. Subtract \frac{1}{2} from 2.5 by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{2}{-1}$$

Divide 2 by -1.

$$x=-2$$

The equation is now solved.

$$x=-3 x=-2$$

Steps for Completing the Square

Subtract 0.5x^{2} from both sides.

$$-x-5-0.5x^{2}=1.5x-2$$

Subtract 1.5x from both sides.

$$-x-5-0.5x^{2}-1.5x=-2$$

Add 5 to both sides.

$$-x-0.5x^{2}-1.5x=-2+5$$

Add -2 and 5 to get 3.

$$-x-0.5x^{2}-1.5x=3$$

Combine -x and -1.5x to get -2.5x.

$$-2.5x-0.5x^{2}=3$$

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.

$$-0.5x^{2}-2.5x=3$$

Multiply both sides by -2.

$$\frac{-0.5x^{2}-2.5x}{-0.5}=\frac{3}{-0.5}$$

Dividing by -0.5 undoes the multiplication by -0.5.

$$x^{2}+\left(-\frac{2.5}{-0.5}\right)x=\frac{3}{-0.5}$$

Divide -2.5 by -0.5 by multiplying -2.5 by the reciprocal of -0.5.

$$x^{2}+5x=\frac{3}{-0.5}$$

Divide 3 by -0.5 by multiplying 3 by the reciprocal of -0.5.

$$x^{2}+5x=-6$$

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.

$$x^{2}+5x+\left(\frac{5}{2}\right)^{2}=-6+\left(\frac{5}{2}\right)^{2}$$

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

$$x^{2}+5x+\frac{25}{4}=-6+\frac{25}{4}$$

Add -6 to \frac{25}{4}.

$$x^{2}+5x+\frac{25}{4}=0.25$$

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}.

$$\left(x+\frac{5}{2}\right)^{2}=0.25$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{2}\right)^{2}}=\sqrt{0.25}$$

Simplify.

$$x+\frac{5}{2}=\frac{1}{2} x+\frac{5}{2}=-\frac{1}{2}$$

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

$$x=-2 x=-3$$