Solve left(x+1right)^2+left(0.75x-7div4right)^2=4

$${ \left(x+1 \right) }^{ 2 } + { \left(0.75x-7 \div 4 \right) }^{ 2 } =4$$

Solve for x

$$x=0.2$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

$$x^{2}+2x+1+\left(0.75x-\frac{7}{4}\right)^{2}=4$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(0.75x-\frac{7}{4}\right)^{2}.

$$x^{2}+2x+1+0.5625x^{2}-\frac{21}{8}x+\frac{49}{16}=4$$

Combine x^{2} and 0.5625x^{2} to get 1.5625x^{2}.

$$1.5625x^{2}+2x+1-\frac{21}{8}x+\frac{49}{16}=4$$

Combine 2x and -\frac{21}{8}x to get -\frac{5}{8}x.

$$1.5625x^{2}-\frac{5}{8}x+1+\frac{49}{16}=4$$

Add 1 and \frac{49}{16} to get \frac{65}{16}.

$$1.5625x^{2}-\frac{5}{8}x+\frac{65}{16}=4$$

Subtract 4 from both sides.

$$1.5625x^{2}-\frac{5}{8}x+\frac{65}{16}-4=0$$

Subtract 4 from \frac{65}{16} to get \frac{1}{16}.

$$1.5625x^{2}-\frac{5}{8}x+\frac{1}{16}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1.5625 for a, -\frac{5}{8} for b, and \frac{1}{16} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-\frac{5}{8}\right)±\sqrt{\left(-\frac{5}{8}\right)^{2}-4\times 1.5625\times \frac{1}{16}}}{2\times 1.5625}$$

Square -\frac{5}{8} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\left(-\frac{5}{8}\right)±\sqrt{\frac{25}{64}-4\times 1.5625\times \frac{1}{16}}}{2\times 1.5625}$$

Multiply -4 times 1.5625.

$$x=\frac{-\left(-\frac{5}{8}\right)±\sqrt{\frac{25}{64}-6.25\times \frac{1}{16}}}{2\times 1.5625}$$

Multiply -6.25 times \frac{1}{16} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\left(-\frac{5}{8}\right)±\sqrt{\frac{25-25}{64}}}{2\times 1.5625}$$

Add \frac{25}{64} to -\frac{25}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\left(-\frac{5}{8}\right)±\sqrt{0}}{2\times 1.5625}$$

Take the square root of 0.

$$x=-\frac{-\frac{5}{8}}{2\times 1.5625}$$

The opposite of -\frac{5}{8} is \frac{5}{8}.

$$x=\frac{\frac{5}{8}}{2\times 1.5625}$$

Multiply 2 times 1.5625.

$$x=\frac{\frac{5}{8}}{3.125}$$

Divide \frac{5}{8} by 3.125 by multiplying \frac{5}{8} by the reciprocal of 3.125.

$$x=\frac{1}{5}$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

$$x^{2}+2x+1+\left(0.75x-\frac{7}{4}\right)^{2}=4$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(0.75x-\frac{7}{4}\right)^{2}.

$$x^{2}+2x+1+0.5625x^{2}-\frac{21}{8}x+\frac{49}{16}=4$$

Combine x^{2} and 0.5625x^{2} to get 1.5625x^{2}.

$$1.5625x^{2}+2x+1-\frac{21}{8}x+\frac{49}{16}=4$$

Combine 2x and -\frac{21}{8}x to get -\frac{5}{8}x.

$$1.5625x^{2}-\frac{5}{8}x+1+\frac{49}{16}=4$$

Add 1 and \frac{49}{16} to get \frac{65}{16}.

$$1.5625x^{2}-\frac{5}{8}x+\frac{65}{16}=4$$

Subtract \frac{65}{16} from both sides.

$$1.5625x^{2}-\frac{5}{8}x=4-\frac{65}{16}$$

Subtract \frac{65}{16} from 4 to get -\frac{1}{16}.

$$1.5625x^{2}-\frac{5}{8}x=-\frac{1}{16}$$

Divide both sides of the equation by 1.5625, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{1.5625x^{2}-\frac{5}{8}x}{1.5625}=-\frac{\frac{1}{16}}{1.5625}$$

Dividing by 1.5625 undoes the multiplication by 1.5625.

$$x^{2}+\left(-\frac{\frac{5}{8}}{1.5625}\right)x=-\frac{\frac{1}{16}}{1.5625}$$

Divide -\frac{5}{8} by 1.5625 by multiplying -\frac{5}{8} by the reciprocal of 1.5625.

$$x^{2}-\frac{2}{5}x=-\frac{\frac{1}{16}}{1.5625}$$

Divide -\frac{1}{16} by 1.5625 by multiplying -\frac{1}{16} by the reciprocal of 1.5625.

$$x^{2}-\frac{2}{5}x=-\frac{1}{25}$$

Divide -\frac{2}{5}, the coefficient of the x term, by 2 to get -\frac{1}{5}. Then add the square of -\frac{1}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{2}{5}x+\left(-\frac{1}{5}\right)^{2}=-\frac{1}{25}+\left(-\frac{1}{5}\right)^{2}$$

Square -\frac{1}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{2}{5}x+\frac{1}{25}=\frac{-1+1}{25}$$

Add -\frac{1}{25} to \frac{1}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{2}{5}x+\frac{1}{25}=0$$

Factor x^{2}-\frac{2}{5}x+\frac{1}{25}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{5}\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{5}\right)^{2}}=\sqrt{0}$$

Simplify.

$$x-\frac{1}{5}=0 x-\frac{1}{5}=0$$

Add \frac{1}{5} to both sides of the equation.

$$x=\frac{1}{5} x=\frac{1}{5}$$

The equation is now solved. Solutions are the same.

$$x=\frac{1}{5}$$