Solve (x+5)(2-3x)=x(x+5)

$$(x+5)(2-3x)=x(x+5)$$

Solve for x

$$x=-5$$

$$x=\frac{1}{2}=0.5$$

Steps Using Factoring By Grouping

Use the distributive property to multiply x+5 by 2-3x and combine like terms.

$$-13x-3x^{2}+10=x\left(x+5\right)$$

Use the distributive property to multiply x by x+5.

$$-13x-3x^{2}+10=x^{2}+5x$$

Subtract x^{2} from both sides.

$$-13x-3x^{2}+10-x^{2}=5x$$

Combine -3x^{2} and -x^{2} to get -4x^{2}.

$$-13x-4x^{2}+10=5x$$

Subtract 5x from both sides.

$$-13x-4x^{2}+10-5x=0$$

Combine -13x and -5x to get -18x.

$$-18x-4x^{2}+10=0$$

Divide both sides by 2.

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

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

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -2x^{2}+ax+bx+5. To find a and b, set up a system to be solved.

$$a+b=-9 ab=-2\times 5=-10$$

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 -10.

$$1,-10 2,-5$$

Calculate the sum for each pair.

$$1-10=-9 2-5=-3$$

The solution is the pair that gives sum -9.

$$a=1 b=-10$$

Rewrite -2x^{2}-9x+5 as \left(-2x^{2}+x\right)+\left(-10x+5\right).

$$\left(-2x^{2}+x\right)+\left(-10x+5\right)$$

Factor out -x in the first and -5 in the second group.

$$-x\left(2x-1\right)-5\left(2x-1\right)$$

Factor out common term 2x-1 by using distributive property.

$$\left(2x-1\right)\left(-x-5\right)$$

To find equation solutions, solve 2x-1=0 and -x-5=0.

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

Steps Using the Quadratic Formula

Use the distributive property to multiply x+5 by 2-3x and combine like terms.

$$-13x-3x^{2}+10=x\left(x+5\right)$$

Use the distributive property to multiply x by x+5.

$$-13x-3x^{2}+10=x^{2}+5x$$

Subtract x^{2} from both sides.

$$-13x-3x^{2}+10-x^{2}=5x$$

Combine -3x^{2} and -x^{2} to get -4x^{2}.

$$-13x-4x^{2}+10=5x$$

Subtract 5x from both sides.

$$-13x-4x^{2}+10-5x=0$$

Combine -13x and -5x to get -18x.

$$-18x-4x^{2}+10=0$$

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.

$$-4x^{2}-18x+10=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -4 for a, -18 for b, and 10 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-18\right)±\sqrt{\left(-18\right)^{2}-4\left(-4\right)\times 10}}{2\left(-4\right)}$$

Square -18.

$$x=\frac{-\left(-18\right)±\sqrt{324-4\left(-4\right)\times 10}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$x=\frac{-\left(-18\right)±\sqrt{324+16\times 10}}{2\left(-4\right)}$$

Multiply 16 times 10.

$$x=\frac{-\left(-18\right)±\sqrt{324+160}}{2\left(-4\right)}$$

Add 324 to 160.

$$x=\frac{-\left(-18\right)±\sqrt{484}}{2\left(-4\right)}$$

Take the square root of 484.

$$x=\frac{-\left(-18\right)±22}{2\left(-4\right)}$$

The opposite of -18 is 18.

$$x=\frac{18±22}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{18±22}{-8}$$

Now solve the equation x=\frac{18±22}{-8} when ± is plus. Add 18 to 22.

$$x=\frac{40}{-8}$$

Divide 40 by -8.

$$x=-5$$

Now solve the equation x=\frac{18±22}{-8} when ± is minus. Subtract 22 from 18.

$$x=-\frac{4}{-8}$$

Reduce the fraction \frac{-4}{-8} to lowest terms by extracting and canceling out 4.

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

The equation is now solved.

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

Steps for Completing the Square

Use the distributive property to multiply x+5 by 2-3x and combine like terms.

$$-13x-3x^{2}+10=x\left(x+5\right)$$

Use the distributive property to multiply x by x+5.

$$-13x-3x^{2}+10=x^{2}+5x$$

Subtract x^{2} from both sides.

$$-13x-3x^{2}+10-x^{2}=5x$$

Combine -3x^{2} and -x^{2} to get -4x^{2}.

$$-13x-4x^{2}+10=5x$$

Subtract 5x from both sides.

$$-13x-4x^{2}+10-5x=0$$

Combine -13x and -5x to get -18x.

$$-18x-4x^{2}+10=0$$

Subtract 10 from both sides. Anything subtracted from zero gives its negation.

$$-18x-4x^{2}=-10$$

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.

$$-4x^{2}-18x=-10$$

Divide both sides by -4.

$$\frac{-4x^{2}-18x}{-4}=-\frac{10}{-4}$$

Dividing by -4 undoes the multiplication by -4.

$$x^{2}+\left(-\frac{18}{-4}\right)x=-\frac{10}{-4}$$

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

$$x^{2}+\frac{9}{2}x=-\frac{10}{-4}$$

Reduce the fraction \frac{-10}{-4} to lowest terms by extracting and canceling out 2.

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

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.

$$x^{2}+\frac{9}{2}x+\left(\frac{9}{4}\right)^{2}=\frac{5}{2}+\left(\frac{9}{4}\right)^{2}$$

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

$$x^{2}+\frac{9}{2}x+\frac{81}{16}=\frac{5}{2}+\frac{81}{16}$$

Add \frac{5}{2} to \frac{81}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{9}{2}x+\frac{81}{16}=\frac{121}{16}$$

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}.

$$\left(x+\frac{9}{4}\right)^{2}=\frac{121}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{9}{4}\right)^{2}}=\sqrt{\frac{121}{16}}$$

Simplify.

$$x+\frac{9}{4}=\frac{11}{4} x+\frac{9}{4}=-\frac{11}{4}$$

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

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