Solve (2x-1)(5+x)=(0,5x+4)(x-3)

$$( 2 x - 1 ) ( 5 + x ) = ( 0,5 x + 4 ) ( x - 3 )$$

Solve for x

$$x=-\frac{7}{3}\approx -2,333333333$$

$$x=-2$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 2x-1 by 5+x and combine like terms.

$$9x+2x^{2}-5=\left(0,5x+4\right)\left(x-3\right)$$

Use the distributive property to multiply 0,5x+4 by x-3 and combine like terms.

$$9x+2x^{2}-5=0,5x^{2}+2,5x-12$$

Subtract 0,5x^{2} from both sides.

$$9x+2x^{2}-5-0,5x^{2}=2,5x-12$$

Combine 2x^{2} and -0,5x^{2} to get 1,5x^{2}.

$$9x+1,5x^{2}-5=2,5x-12$$

Subtract 2,5x from both sides.

$$9x+1,5x^{2}-5-2,5x=-12$$

Combine 9x and -2,5x to get 6,5x.

$$6,5x+1,5x^{2}-5=-12$$

Add 12 to both sides.

$$6,5x+1,5x^{2}-5+12=0$$

Add -5 and 12 to get 7.

$$6,5x+1,5x^{2}+7=0$$

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.

$$1,5x^{2}+6,5x+7=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1,5 for a, 6,5 for b, and 7 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-6,5±\sqrt{6,5^{2}-4\times 1,5\times 7}}{2\times 1,5}$$

Square 6,5 by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-6,5±\sqrt{42,25-4\times 1,5\times 7}}{2\times 1,5}$$

Multiply -4 times 1,5.

$$x=\frac{-6,5±\sqrt{42,25-6\times 7}}{2\times 1,5}$$

Multiply -6 times 7.

$$x=\frac{-6,5±\sqrt{42,25-42}}{2\times 1,5}$$

Add 42,25 to -42.

$$x=\frac{-6,5±\sqrt{0,25}}{2\times 1,5}$$

Take the square root of 0,25.

$$x=\frac{-6,5±\frac{1}{2}}{2\times 1,5}$$

Multiply 2 times 1,5.

$$x=\frac{-6,5±\frac{1}{2}}{3}$$

Now solve the equation x=\frac{-6,5±\frac{1}{2}}{3} when ± is plus. Add -6,5 to \frac{1}{2} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=-\frac{6}{3}$$

Divide -6 by 3.

$$x=-2$$

Now solve the equation x=\frac{-6,5±\frac{1}{2}}{3} when ± is minus. Subtract \frac{1}{2} from -6,5 by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$x=-\frac{7}{3}$$

The equation is now solved.

$$x=-2 x=-\frac{7}{3}$$

Steps for Completing the Square

Use the distributive property to multiply 2x-1 by 5+x and combine like terms.

$$9x+2x^{2}-5=\left(0,5x+4\right)\left(x-3\right)$$

Use the distributive property to multiply 0,5x+4 by x-3 and combine like terms.

$$9x+2x^{2}-5=0,5x^{2}+2,5x-12$$

Subtract 0,5x^{2} from both sides.

$$9x+2x^{2}-5-0,5x^{2}=2,5x-12$$

Combine 2x^{2} and -0,5x^{2} to get 1,5x^{2}.

$$9x+1,5x^{2}-5=2,5x-12$$

Subtract 2,5x from both sides.

$$9x+1,5x^{2}-5-2,5x=-12$$

Combine 9x and -2,5x to get 6,5x.

$$6,5x+1,5x^{2}-5=-12$$

Add 5 to both sides.

$$6,5x+1,5x^{2}=-12+5$$

Add -12 and 5 to get -7.

$$6,5x+1,5x^{2}=-7$$

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.

$$1,5x^{2}+6,5x=-7$$

Divide both sides of the equation by 1,5, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{1,5x^{2}+6,5x}{1,5}=-\frac{7}{1,5}$$

Dividing by 1,5 undoes the multiplication by 1,5.

$$x^{2}+\frac{6,5}{1,5}x=-\frac{7}{1,5}$$

Divide 6,5 by 1,5 by multiplying 6,5 by the reciprocal of 1,5.

$$x^{2}+\frac{13}{3}x=-\frac{7}{1,5}$$

Divide -7 by 1,5 by multiplying -7 by the reciprocal of 1,5.

$$x^{2}+\frac{13}{3}x=-\frac{14}{3}$$

Divide \frac{13}{3}, the coefficient of the x term, by 2 to get \frac{13}{6}. Then add the square of \frac{13}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{13}{3}x+\frac{13}{6}^{2}=-\frac{14}{3}+\frac{13}{6}^{2}$$

Square \frac{13}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{13}{3}x+\frac{169}{36}=-\frac{14}{3}+\frac{169}{36}$$

Add -\frac{14}{3} to \frac{169}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{13}{3}x+\frac{169}{36}=\frac{1}{36}$$

Factor x^{2}+\frac{13}{3}x+\frac{169}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{13}{6}\right)^{2}=\frac{1}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{13}{6}\right)^{2}}=\sqrt{\frac{1}{36}}$$

Simplify.

$$x+\frac{13}{6}=\frac{1}{6} x+\frac{13}{6}=-\frac{1}{6}$$

Subtract \frac{13}{6} from both sides of the equation.

$$x=-2 x=-\frac{7}{3}$$