Solve (2x+1)*(x+5)=8*5

$$( 2 x + 1 ) \cdot ( x + 5 ) = 8 \cdot 5$$

Solve for x

$$x = \frac{\sqrt{401} - 11}{4} \approx 2.256246099$$

$$x=\frac{-\sqrt{401}-11}{4}\approx -7.756246099$$

Steps Using the Quadratic Formula

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

$$2x^{2}+11x+5=8\times 5$$

Multiply 8 and 5 to get 40.

$$2x^{2}+11x+5=40$$

Subtract 40 from both sides.

$$2x^{2}+11x+5-40=0$$

Subtract 40 from 5 to get -35.

$$2x^{2}+11x-35=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 11 for b, and -35 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-11±\sqrt{11^{2}-4\times 2\left(-35\right)}}{2\times 2}$$

Square 11.

$$x=\frac{-11±\sqrt{121-4\times 2\left(-35\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-11±\sqrt{121-8\left(-35\right)}}{2\times 2}$$

Multiply -8 times -35.

$$x=\frac{-11±\sqrt{121+280}}{2\times 2}$$

Add 121 to 280.

$$x=\frac{-11±\sqrt{401}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-11±\sqrt{401}}{4}$$

Now solve the equation x=\frac{-11±\sqrt{401}}{4} when ± is plus. Add -11 to \sqrt{401}.

$$x=\frac{\sqrt{401}-11}{4}$$

Now solve the equation x=\frac{-11±\sqrt{401}}{4} when ± is minus. Subtract \sqrt{401} from -11.

$$x=\frac{-\sqrt{401}-11}{4}$$

The equation is now solved.

$$x=\frac{\sqrt{401}-11}{4} x=\frac{-\sqrt{401}-11}{4}$$

Steps for Completing the Square

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

$$2x^{2}+11x+5=8\times 5$$

Multiply 8 and 5 to get 40.

$$2x^{2}+11x+5=40$$

Subtract 5 from both sides.

$$2x^{2}+11x=40-5$$

Subtract 5 from 40 to get 35.

$$2x^{2}+11x=35$$

Divide both sides by 2.

$$\frac{2x^{2}+11x}{2}=\frac{35}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{11}{2}x=\frac{35}{2}$$

Divide \frac{11}{2}, the coefficient of the x term, by 2 to get \frac{11}{4}. Then add the square of \frac{11}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{11}{2}x+\left(\frac{11}{4}\right)^{2}=\frac{35}{2}+\left(\frac{11}{4}\right)^{2}$$

Square \frac{11}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{11}{2}x+\frac{121}{16}=\frac{35}{2}+\frac{121}{16}$$

Add \frac{35}{2} to \frac{121}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{11}{2}x+\frac{121}{16}=\frac{401}{16}$$

Factor x^{2}+\frac{11}{2}x+\frac{121}{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}.

$$\left(x+\frac{11}{4}\right)^{2}=\frac{401}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{11}{4}\right)^{2}}=\sqrt{\frac{401}{16}}$$

Simplify.

$$x+\frac{11}{4}=\frac{\sqrt{401}}{4} x+\frac{11}{4}=-\frac{\sqrt{401}}{4}$$

Subtract \frac{11}{4} from both sides of the equation.

$$x=\frac{\sqrt{401}-11}{4} x=\frac{-\sqrt{401}-11}{4}$$