Solve -2x^2+2x+9=-5x

$$- 2 x ^ { 2 } + 2 x + 9 = - 5 x$$

Solve for x

$$x=-1$$

$$x = \frac{9}{2} = 4\frac{1}{2} = 4.5$$

Steps Using Factoring By Grouping

Add 5x to both sides.

$$-2x^{2}+2x+9+5x=0$$

Combine 2x and 5x to get 7x.

$$-2x^{2}+7x+9=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -2x^{2}+ax+bx+9. To find a and b, set up a system to be solved.

$$a+b=7 ab=-2\times 9=-18$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -18.

$$-1,18 -2,9 -3,6$$

Calculate the sum for each pair.

$$-1+18=17 -2+9=7 -3+6=3$$

The solution is the pair that gives sum 7.

$$a=9 b=-2$$

Rewrite -2x^{2}+7x+9 as \left(-2x^{2}+9x\right)+\left(-2x+9\right).

$$\left(-2x^{2}+9x\right)+\left(-2x+9\right)$$

Factor out -x in the first and -1 in the second group.

$$-x\left(2x-9\right)-\left(2x-9\right)$$

Factor out common term 2x-9 by using distributive property.

$$\left(2x-9\right)\left(-x-1\right)$$

To find equation solutions, solve 2x-9=0 and -x-1=0.

$$x=\frac{9}{2} x=-1$$

Steps Using the Quadratic Formula

Add 5x to both sides.

$$-2x^{2}+2x+9+5x=0$$

Combine 2x and 5x to get 7x.

$$-2x^{2}+7x+9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, 7 for b, and 9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-7±\sqrt{7^{2}-4\left(-2\right)\times 9}}{2\left(-2\right)}$$

Square 7.

$$x=\frac{-7±\sqrt{49-4\left(-2\right)\times 9}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-7±\sqrt{49+8\times 9}}{2\left(-2\right)}$$

Multiply 8 times 9.

$$x=\frac{-7±\sqrt{49+72}}{2\left(-2\right)}$$

Add 49 to 72.

$$x=\frac{-7±\sqrt{121}}{2\left(-2\right)}$$

Take the square root of 121.

$$x=\frac{-7±11}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{-7±11}{-4}$$

Now solve the equation x=\frac{-7±11}{-4} when ± is plus. Add -7 to 11.

$$x=\frac{4}{-4}$$

Divide 4 by -4.

$$x=-1$$

Now solve the equation x=\frac{-7±11}{-4} when ± is minus. Subtract 11 from -7.

$$x=-\frac{18}{-4}$$

Reduce the fraction \frac{-18}{-4} to lowest terms by extracting and canceling out 2.

$$x=\frac{9}{2}$$

The equation is now solved.

$$x=-1 x=\frac{9}{2}$$

Steps for Completing the Square

Add 5x to both sides.

$$-2x^{2}+2x+9+5x=0$$

Combine 2x and 5x to get 7x.

$$-2x^{2}+7x+9=0$$

Subtract 9 from both sides. Anything subtracted from zero gives its negation.

$$-2x^{2}+7x=-9$$

Divide both sides by -2.

$$\frac{-2x^{2}+7x}{-2}=-\frac{9}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$x^{2}+\frac{7}{-2}x=-\frac{9}{-2}$$

Divide 7 by -2.

$$x^{2}-\frac{7}{2}x=-\frac{9}{-2}$$

Divide -9 by -2.

$$x^{2}-\frac{7}{2}x=\frac{9}{2}$$

Divide -\frac{7}{2}, the coefficient of the x term, by 2 to get -\frac{7}{4}. Then add the square of -\frac{7}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{7}{2}x+\left(-\frac{7}{4}\right)^{2}=\frac{9}{2}+\left(-\frac{7}{4}\right)^{2}$$

Square -\frac{7}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{7}{2}x+\frac{49}{16}=\frac{9}{2}+\frac{49}{16}$$

Add \frac{9}{2} to \frac{49}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{7}{2}x+\frac{49}{16}=\frac{121}{16}$$

Factor x^{2}-\frac{7}{2}x+\frac{49}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{7}{4}\right)^{2}=\frac{121}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{4}\right)^{2}}=\sqrt{\frac{121}{16}}$$

Simplify.

$$x-\frac{7}{4}=\frac{11}{4} x-\frac{7}{4}=-\frac{11}{4}$$

Add \frac{7}{4} to both sides of the equation.

$$x=\frac{9}{2} x=-1$$