Solve (2x+3)*(x+5)=41

$$(2x+3) \times (x+5)=41$$

Solve for x

$$x = \frac{\sqrt{377} - 13}{4} \approx 1.60412196$$

$$x=\frac{-\sqrt{377}-13}{4}\approx -8.10412196$$

Steps Using the Quadratic Formula

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

$$2x^{2}+13x+15=41$$

Subtract 41 from both sides.

$$2x^{2}+13x+15-41=0$$

Subtract 41 from 15 to get -26.

$$2x^{2}+13x-26=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 13 for b, and -26 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-13±\sqrt{13^{2}-4\times 2\left(-26\right)}}{2\times 2}$$

Square 13.

$$x=\frac{-13±\sqrt{169-4\times 2\left(-26\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-13±\sqrt{169-8\left(-26\right)}}{2\times 2}$$

Multiply -8 times -26.

$$x=\frac{-13±\sqrt{169+208}}{2\times 2}$$

Add 169 to 208.

$$x=\frac{-13±\sqrt{377}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-13±\sqrt{377}}{4}$$

Now solve the equation x=\frac{-13±\sqrt{377}}{4} when ± is plus. Add -13 to \sqrt{377}.

$$x=\frac{\sqrt{377}-13}{4}$$

Now solve the equation x=\frac{-13±\sqrt{377}}{4} when ± is minus. Subtract \sqrt{377} from -13.

$$x=\frac{-\sqrt{377}-13}{4}$$

The equation is now solved.

$$x=\frac{\sqrt{377}-13}{4} x=\frac{-\sqrt{377}-13}{4}$$

Steps for Completing the Square

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

$$2x^{2}+13x+15=41$$

Subtract 15 from both sides.

$$2x^{2}+13x=41-15$$

Subtract 15 from 41 to get 26.

$$2x^{2}+13x=26$$

Divide both sides by 2.

$$\frac{2x^{2}+13x}{2}=\frac{26}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{13}{2}x=\frac{26}{2}$$

Divide 26 by 2.

$$x^{2}+\frac{13}{2}x=13$$

Divide \frac{13}{2}, the coefficient of the x term, by 2 to get \frac{13}{4}. Then add the square of \frac{13}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{13}{2}x+\left(\frac{13}{4}\right)^{2}=13+\left(\frac{13}{4}\right)^{2}$$

Square \frac{13}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{13}{2}x+\frac{169}{16}=13+\frac{169}{16}$$

Add 13 to \frac{169}{16}.

$$x^{2}+\frac{13}{2}x+\frac{169}{16}=\frac{377}{16}$$

Factor x^{2}+\frac{13}{2}x+\frac{169}{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}.

$$\left(x+\frac{13}{4}\right)^{2}=\frac{377}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{13}{4}\right)^{2}}=\sqrt{\frac{377}{16}}$$

Simplify.

$$x+\frac{13}{4}=\frac{\sqrt{377}}{4} x+\frac{13}{4}=-\frac{\sqrt{377}}{4}$$

Subtract \frac{13}{4} from both sides of the equation.

$$x=\frac{\sqrt{377}-13}{4} x=\frac{-\sqrt{377}-13}{4}$$