Solve x(3-x)+x(2-x)=-25

$$x ( 3 - x ) + x ( 2 - x ) = - 25$$

Solve for x

$$x = -\frac{5}{2} = -2\frac{1}{2} = -2.5$$

$$x=5$$

Steps Using the Quadratic Formula

Use the distributive property to multiply x by 3-x.

$$3x-x^{2}+x\left(2-x\right)=-25$$

Use the distributive property to multiply x by 2-x.

$$3x-x^{2}+2x-x^{2}=-25$$

Combine 3x and 2x to get 5x.

$$5x-x^{2}-x^{2}=-25$$

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

$$5x-2x^{2}=-25$$

Add 25 to both sides.

$$5x-2x^{2}+25=0$$

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.

$$-2x^{2}+5x+25=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, 5 for b, and 25 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square 5.

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

Multiply -4 times -2.

$$x=\frac{-5±\sqrt{25+8\times 25}}{2\left(-2\right)}$$

Multiply 8 times 25.

$$x=\frac{-5±\sqrt{25+200}}{2\left(-2\right)}$$

Add 25 to 200.

$$x=\frac{-5±\sqrt{225}}{2\left(-2\right)}$$

Take the square root of 225.

$$x=\frac{-5±15}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{-5±15}{-4}$$

Now solve the equation x=\frac{-5±15}{-4} when ± is plus. Add -5 to 15.

$$x=\frac{10}{-4}$$

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

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

Now solve the equation x=\frac{-5±15}{-4} when ± is minus. Subtract 15 from -5.

$$x=-\frac{20}{-4}$$

Divide -20 by -4.

$$x=5$$

The equation is now solved.

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

Steps for Completing the Square

Use the distributive property to multiply x by 3-x.

$$3x-x^{2}+x\left(2-x\right)=-25$$

Use the distributive property to multiply x by 2-x.

$$3x-x^{2}+2x-x^{2}=-25$$

Combine 3x and 2x to get 5x.

$$5x-x^{2}-x^{2}=-25$$

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

$$5x-2x^{2}=-25$$

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.

$$-2x^{2}+5x=-25$$

Divide both sides by -2.

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

Dividing by -2 undoes the multiplication by -2.

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

Divide 5 by -2.

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

Divide -25 by -2.

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

Divide -\frac{5}{2}, the coefficient of the x term, by 2 to get -\frac{5}{4}. Then add the square of -\frac{5}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{5}{2}x+\left(-\frac{5}{4}\right)^{2}=\frac{25}{2}+\left(-\frac{5}{4}\right)^{2}$$

Square -\frac{5}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{5}{2}x+\frac{25}{16}=\frac{25}{2}+\frac{25}{16}$$

Add \frac{25}{2} to \frac{25}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{5}{2}x+\frac{25}{16}=\frac{225}{16}$$

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

$$\left(x-\frac{5}{4}\right)^{2}=\frac{225}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{4}\right)^{2}}=\sqrt{\frac{225}{16}}$$

Simplify.

$$x-\frac{5}{4}=\frac{15}{4} x-\frac{5}{4}=-\frac{15}{4}$$

Add \frac{5}{4} to both sides of the equation.

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