Solve 5x^2-(x+1)^2=1900

$$5 x ^ { 2 } - ( x + 1 ) ^ { 2 } = 1900$$

Solve for x

$$x = \frac{39 \sqrt{5} + 1}{4} \approx 22.051662781$$

$$x=\frac{1-39\sqrt{5}}{4}\approx -21.551662781$$

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}.

$$5x^{2}-\left(x^{2}+2x+1\right)=1900$$

To find the opposite of x^{2}+2x+1, find the opposite of each term.

$$5x^{2}-x^{2}-2x-1=1900$$

Combine 5x^{2} and -x^{2} to get 4x^{2}.

$$4x^{2}-2x-1=1900$$

Subtract 1900 from both sides.

$$4x^{2}-2x-1-1900=0$$

Subtract 1900 from -1 to get -1901.

$$4x^{2}-2x-1901=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -2 for b, and -1901 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times 4\left(-1901\right)}}{2\times 4}$$

Square -2.

$$x=\frac{-\left(-2\right)±\sqrt{4-4\times 4\left(-1901\right)}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-2\right)±\sqrt{4-16\left(-1901\right)}}{2\times 4}$$

Multiply -16 times -1901.

$$x=\frac{-\left(-2\right)±\sqrt{4+30416}}{2\times 4}$$

Add 4 to 30416.

$$x=\frac{-\left(-2\right)±\sqrt{30420}}{2\times 4}$$

Take the square root of 30420.

$$x=\frac{-\left(-2\right)±78\sqrt{5}}{2\times 4}$$

The opposite of -2 is 2.

$$x=\frac{2±78\sqrt{5}}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{2±78\sqrt{5}}{8}$$

Now solve the equation x=\frac{2±78\sqrt{5}}{8} when ± is plus. Add 2 to 78\sqrt{5}.

$$x=\frac{78\sqrt{5}+2}{8}$$

Divide 2+78\sqrt{5} by 8.

$$x=\frac{39\sqrt{5}+1}{4}$$

Now solve the equation x=\frac{2±78\sqrt{5}}{8} when ± is minus. Subtract 78\sqrt{5} from 2.

$$x=\frac{2-78\sqrt{5}}{8}$$

Divide 2-78\sqrt{5} by 8.

$$x=\frac{1-39\sqrt{5}}{4}$$

The equation is now solved.

$$x=\frac{39\sqrt{5}+1}{4} x=\frac{1-39\sqrt{5}}{4}$$

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}.

$$5x^{2}-\left(x^{2}+2x+1\right)=1900$$

To find the opposite of x^{2}+2x+1, find the opposite of each term.

$$5x^{2}-x^{2}-2x-1=1900$$

Combine 5x^{2} and -x^{2} to get 4x^{2}.

$$4x^{2}-2x-1=1900$$

Add 1 to both sides.

$$4x^{2}-2x=1900+1$$

Add 1900 and 1 to get 1901.

$$4x^{2}-2x=1901$$

Divide both sides by 4.

$$\frac{4x^{2}-2x}{4}=\frac{1901}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\left(-\frac{2}{4}\right)x=\frac{1901}{4}$$

Reduce the fraction \frac{-2}{4} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{1}{2}x=\frac{1901}{4}$$

Divide -\frac{1}{2}, the coefficient of the x term, by 2 to get -\frac{1}{4}. Then add the square of -\frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1}{2}x+\left(-\frac{1}{4}\right)^{2}=\frac{1901}{4}+\left(-\frac{1}{4}\right)^{2}$$

Square -\frac{1}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=\frac{1901}{4}+\frac{1}{16}$$

Add \frac{1901}{4} to \frac{1}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=\frac{7605}{16}$$

Factor x^{2}-\frac{1}{2}x+\frac{1}{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}.

$$\left(x-\frac{1}{4}\right)^{2}=\frac{7605}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{4}\right)^{2}}=\sqrt{\frac{7605}{16}}$$

Simplify.

$$x-\frac{1}{4}=\frac{39\sqrt{5}}{4} x-\frac{1}{4}=-\frac{39\sqrt{5}}{4}$$

Add \frac{1}{4} to both sides of the equation.

$$x=\frac{39\sqrt{5}+1}{4} x=\frac{1-39\sqrt{5}}{4}$$