Solve 5-x=2x^2/9

$$5 - x = \frac { 2 x ^ { 2 } } { 9 }$$

Solve for x

$$x = -\frac{15}{2} = -7\frac{1}{2} = -7.5$$

$$x=3$$

Steps Using Factoring By Grouping

Multiply both sides of the equation by 9.

$$45-9x=2x^{2}$$

Subtract 2x^{2} from both sides.

$$45-9x-2x^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-2x^{2}-9x+45=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -2x^{2}+ax+bx+45. To find a and b, set up a system to be solved.

$$a+b=-9 ab=-2\times 45=-90$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -90.

$$1,-90 2,-45 3,-30 5,-18 6,-15 9,-10$$

Calculate the sum for each pair.

$$1-90=-89 2-45=-43 3-30=-27 5-18=-13 6-15=-9 9-10=-1$$

The solution is the pair that gives sum -9.

$$a=6 b=-15$$

Rewrite -2x^{2}-9x+45 as \left(-2x^{2}+6x\right)+\left(-15x+45\right).

$$\left(-2x^{2}+6x\right)+\left(-15x+45\right)$$

Factor out 2x in the first and 15 in the second group.

$$2x\left(-x+3\right)+15\left(-x+3\right)$$

Factor out common term -x+3 by using distributive property.

$$\left(-x+3\right)\left(2x+15\right)$$

To find equation solutions, solve -x+3=0 and 2x+15=0.

$$x=3 x=-\frac{15}{2}$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 9.

$$45-9x=2x^{2}$$

Subtract 2x^{2} from both sides.

$$45-9x-2x^{2}=0$$

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.

$$-2x^{2}-9x+45=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, -9 for b, and 45 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\left(-2\right)\times 45}}{2\left(-2\right)}$$

Square -9.

$$x=\frac{-\left(-9\right)±\sqrt{81-4\left(-2\right)\times 45}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-\left(-9\right)±\sqrt{81+8\times 45}}{2\left(-2\right)}$$

Multiply 8 times 45.

$$x=\frac{-\left(-9\right)±\sqrt{81+360}}{2\left(-2\right)}$$

Add 81 to 360.

$$x=\frac{-\left(-9\right)±\sqrt{441}}{2\left(-2\right)}$$

Take the square root of 441.

$$x=\frac{-\left(-9\right)±21}{2\left(-2\right)}$$

The opposite of -9 is 9.

$$x=\frac{9±21}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{9±21}{-4}$$

Now solve the equation x=\frac{9±21}{-4} when ± is plus. Add 9 to 21.

$$x=\frac{30}{-4}$$

Reduce the fraction \frac{30}{-4} to lowest terms by extracting and canceling out 2.

$$x=-\frac{15}{2}$$

Now solve the equation x=\frac{9±21}{-4} when ± is minus. Subtract 21 from 9.

$$x=-\frac{12}{-4}$$

Divide -12 by -4.

$$x=3$$

The equation is now solved.

$$x=-\frac{15}{2} x=3$$

Steps for Completing the Square

Multiply both sides of the equation by 9.

$$45-9x=2x^{2}$$

Subtract 2x^{2} from both sides.

$$45-9x-2x^{2}=0$$

Subtract 45 from both sides. Anything subtracted from zero gives its negation.

$$-9x-2x^{2}=-45$$

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.

$$-2x^{2}-9x=-45$$

Divide both sides by -2.

$$\frac{-2x^{2}-9x}{-2}=-\frac{45}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$x^{2}+\left(-\frac{9}{-2}\right)x=-\frac{45}{-2}$$

Divide -9 by -2.

$$x^{2}+\frac{9}{2}x=-\frac{45}{-2}$$

Divide -45 by -2.

$$x^{2}+\frac{9}{2}x=\frac{45}{2}$$

Divide \frac{9}{2}, the coefficient of the x term, by 2 to get \frac{9}{4}. Then add the square of \frac{9}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{9}{2}x+\left(\frac{9}{4}\right)^{2}=\frac{45}{2}+\left(\frac{9}{4}\right)^{2}$$

Square \frac{9}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{9}{2}x+\frac{81}{16}=\frac{45}{2}+\frac{81}{16}$$

Add \frac{45}{2} to \frac{81}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{9}{2}x+\frac{81}{16}=\frac{441}{16}$$

Factor x^{2}+\frac{9}{2}x+\frac{81}{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}.

$$\left(x+\frac{9}{4}\right)^{2}=\frac{441}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{9}{4}\right)^{2}}=\sqrt{\frac{441}{16}}$$

Simplify.

$$x+\frac{9}{4}=\frac{21}{4} x+\frac{9}{4}=-\frac{21}{4}$$

Subtract \frac{9}{4} from both sides of the equation.

$$x=3 x=-\frac{15}{2}$$