Solve 5x*(x+5)=1500

$$5 x \cdot ( x + 5 ) = 1500$$

Solve for x

$$x=-20$$

$$x=15$$

Steps Using the Quadratic Formula

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

$$5x^{2}+25x=1500$$

Subtract 1500 from both sides.

$$5x^{2}+25x-1500=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, 25 for b, and -1500 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-25±\sqrt{25^{2}-4\times 5\left(-1500\right)}}{2\times 5}$$

Square 25.

$$x=\frac{-25±\sqrt{625-4\times 5\left(-1500\right)}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-25±\sqrt{625-20\left(-1500\right)}}{2\times 5}$$

Multiply -20 times -1500.

$$x=\frac{-25±\sqrt{625+30000}}{2\times 5}$$

Add 625 to 30000.

$$x=\frac{-25±\sqrt{30625}}{2\times 5}$$

Take the square root of 30625.

$$x=\frac{-25±175}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{-25±175}{10}$$

Now solve the equation x=\frac{-25±175}{10} when ± is plus. Add -25 to 175.

$$x=\frac{150}{10}$$

Divide 150 by 10.

$$x=15$$

Now solve the equation x=\frac{-25±175}{10} when ± is minus. Subtract 175 from -25.

$$x=-\frac{200}{10}$$

Divide -200 by 10.

$$x=-20$$

The equation is now solved.

$$x=15 x=-20$$

Steps for Completing the Square

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

$$5x^{2}+25x=1500$$

Divide both sides by 5.

$$\frac{5x^{2}+25x}{5}=\frac{1500}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$x^{2}+\frac{25}{5}x=\frac{1500}{5}$$

Divide 25 by 5.

$$x^{2}+5x=\frac{1500}{5}$$

Divide 1500 by 5.

$$x^{2}+5x=300$$

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

$$x^{2}+5x+\left(\frac{5}{2}\right)^{2}=300+\left(\frac{5}{2}\right)^{2}$$

Square \frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+5x+\frac{25}{4}=300+\frac{25}{4}$$

Add 300 to \frac{25}{4}.

$$x^{2}+5x+\frac{25}{4}=\frac{1225}{4}$$

Factor x^{2}+5x+\frac{25}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{5}{2}\right)^{2}=\frac{1225}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{2}\right)^{2}}=\sqrt{\frac{1225}{4}}$$

Simplify.

$$x+\frac{5}{2}=\frac{35}{2} x+\frac{5}{2}=-\frac{35}{2}$$

Subtract \frac{5}{2} from both sides of the equation.

$$x=15 x=-20$$