Solve p^2+2p(0.3)+(0.3)^2=1

$$p ^ { 2 } + 2 p ( 0.3 ) + ( 0.3 ) ^ { 2 } = 1$$

Solve for p

$$p=-1.3$$

$$p=0.7$$

Steps Using the Quadratic Formula

Multiply 2 and 0.3 to get 0.6.

$$p^{2}+0.6p+0.3^{2}=1$$

Calculate 0.3 to the power of 2 and get 0.09.

$$p^{2}+0.6p+0.09=1$$

Subtract 1 from both sides.

$$p^{2}+0.6p+0.09-1=0$$

Subtract 1 from 0.09 to get -0.91.

$$p^{2}+0.6p-0.91=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0.6 for b, and -0.91 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$p=\frac{-0.6±\sqrt{0.6^{2}-4\left(-0.91\right)}}{2}$$

Square 0.6 by squaring both the numerator and the denominator of the fraction.

$$p=\frac{-0.6±\sqrt{0.36-4\left(-0.91\right)}}{2}$$

Multiply -4 times -0.91.

$$p=\frac{-0.6±\sqrt{\frac{9+91}{25}}}{2}$$

Add 0.36 to 3.64 by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$p=\frac{-0.6±\sqrt{4}}{2}$$

Take the square root of 4.

$$p=\frac{-0.6±2}{2}$$

Now solve the equation p=\frac{-0.6±2}{2} when ± is plus. Add -0.6 to 2.

$$p=\frac{1.4}{2}$$

Divide 1.4 by 2.

$$p=0.7$$

Now solve the equation p=\frac{-0.6±2}{2} when ± is minus. Subtract 2 from -0.6.

$$p=-\frac{2.6}{2}$$

Divide -2.6 by 2.

$$p=-1.3$$

The equation is now solved.

$$p=0.7 p=-1.3$$

Steps for Completing the Square

Multiply 2 and 0.3 to get 0.6.

$$p^{2}+0.6p+0.3^{2}=1$$

Calculate 0.3 to the power of 2 and get 0.09.

$$p^{2}+0.6p+0.09=1$$

Factor p^{2}+0.6p+0.09. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(p+0.3\right)^{2}=1$$

Take the square root of both sides of the equation.

$$\sqrt{\left(p+0.3\right)^{2}}=\sqrt{1}$$

Simplify.

$$p+0.3=1 p+0.3=-1$$

Subtract 0.3 from both sides of the equation.

$$p=0.7 p=-1.3$$