Solve 0=left(x+1.5right)^2-25

$$0= { \left(x+1.5 \right) }^{ 2 } -25$$

Solve for x

$$x=3.5$$

$$x=-6.5$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1.5\right)^{2}.

$$0=x^{2}+3x+2.25-25$$

Subtract 25 from 2.25 to get -22.75.

$$0=x^{2}+3x-22.75$$

Swap sides so that all variable terms are on the left hand side.

$$x^{2}+3x-22.75=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 3 for b, and -22.75 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-3±\sqrt{3^{2}-4\left(-22.75\right)}}{2}$$

Square 3.

$$x=\frac{-3±\sqrt{9-4\left(-22.75\right)}}{2}$$

Multiply -4 times -22.75.

$$x=\frac{-3±\sqrt{9+91}}{2}$$

Add 9 to 91.

$$x=\frac{-3±\sqrt{100}}{2}$$

Take the square root of 100.

$$x=\frac{-3±10}{2}$$

Now solve the equation x=\frac{-3±10}{2} when ± is plus. Add -3 to 10.

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

Now solve the equation x=\frac{-3±10}{2} when ± is minus. Subtract 10 from -3.

$$x=-\frac{13}{2}$$

The equation is now solved.

$$x=\frac{7}{2} x=-\frac{13}{2}$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1.5\right)^{2}.

$$0=x^{2}+3x+2.25-25$$

Subtract 25 from 2.25 to get -22.75.

$$0=x^{2}+3x-22.75$$

Swap sides so that all variable terms are on the left hand side.

$$x^{2}+3x-22.75=0$$

Add 22.75 to both sides. Anything plus zero gives itself.

$$x^{2}+3x=22.75$$

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

$$x^{2}+3x+\left(\frac{3}{2}\right)^{2}=22.75+\left(\frac{3}{2}\right)^{2}$$

Square \frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+3x+\frac{9}{4}=\frac{91+9}{4}$$

Add 22.75 to \frac{9}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+3x+\frac{9}{4}=25$$

Factor x^{2}+3x+\frac{9}{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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x+\frac{3}{2}=5 x+\frac{3}{2}=-5$$

Subtract \frac{3}{2} from both sides of the equation.

$$x=\frac{7}{2} x=-\frac{13}{2}$$