Solve left(4xright)^2-4x-5=10

$${ \left(4x \right) }^{ 2 } -4x-5 = 10$$

Solve for x

$$x = \frac{\sqrt{61} + 1}{8} \approx 1.101281209$$

$$x=\frac{1-\sqrt{61}}{8}\approx -0.851281209$$

Steps Using the Quadratic Formula

Expand \left(4x\right)^{2}.

$$4^{2}x^{2}-4x-5=10$$

Calculate 4 to the power of 2 and get 16.

$$16x^{2}-4x-5=10$$

Subtract 10 from both sides.

$$16x^{2}-4x-5-10=0$$

Subtract 10 from -5 to get -15.

$$16x^{2}-4x-15=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 16 for a, -4 for b, and -15 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 16\left(-15\right)}}{2\times 16}$$

Square -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\times 16\left(-15\right)}}{2\times 16}$$

Multiply -4 times 16.

$$x=\frac{-\left(-4\right)±\sqrt{16-64\left(-15\right)}}{2\times 16}$$

Multiply -64 times -15.

$$x=\frac{-\left(-4\right)±\sqrt{16+960}}{2\times 16}$$

Add 16 to 960.

$$x=\frac{-\left(-4\right)±\sqrt{976}}{2\times 16}$$

Take the square root of 976.

$$x=\frac{-\left(-4\right)±4\sqrt{61}}{2\times 16}$$

The opposite of -4 is 4.

$$x=\frac{4±4\sqrt{61}}{2\times 16}$$

Multiply 2 times 16.

$$x=\frac{4±4\sqrt{61}}{32}$$

Now solve the equation x=\frac{4±4\sqrt{61}}{32} when ± is plus. Add 4 to 4\sqrt{61}.

$$x=\frac{4\sqrt{61}+4}{32}$$

Divide 4+4\sqrt{61} by 32.

$$x=\frac{\sqrt{61}+1}{8}$$

Now solve the equation x=\frac{4±4\sqrt{61}}{32} when ± is minus. Subtract 4\sqrt{61} from 4.

$$x=\frac{4-4\sqrt{61}}{32}$$

Divide 4-4\sqrt{61} by 32.

$$x=\frac{1-\sqrt{61}}{8}$$

The equation is now solved.

$$x=\frac{\sqrt{61}+1}{8} x=\frac{1-\sqrt{61}}{8}$$

Steps for Completing the Square

Expand \left(4x\right)^{2}.

$$4^{2}x^{2}-4x-5=10$$

Calculate 4 to the power of 2 and get 16.

$$16x^{2}-4x-5=10$$

Add 5 to both sides.

$$16x^{2}-4x=10+5$$

Add 10 and 5 to get 15.

$$16x^{2}-4x=15$$

Divide both sides by 16.

$$\frac{16x^{2}-4x}{16}=\frac{15}{16}$$

Dividing by 16 undoes the multiplication by 16.

$$x^{2}+\left(-\frac{4}{16}\right)x=\frac{15}{16}$$

Reduce the fraction \frac{-4}{16} to lowest terms by extracting and canceling out 4.

$$x^{2}-\frac{1}{4}x=\frac{15}{16}$$

Divide -\frac{1}{4}, the coefficient of the x term, by 2 to get -\frac{1}{8}. Then add the square of -\frac{1}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1}{4}x+\left(-\frac{1}{8}\right)^{2}=\frac{15}{16}+\left(-\frac{1}{8}\right)^{2}$$

Square -\frac{1}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{4}x+\frac{1}{64}=\frac{15}{16}+\frac{1}{64}$$

Add \frac{15}{16} to \frac{1}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{1}{4}x+\frac{1}{64}=\frac{61}{64}$$

Factor x^{2}-\frac{1}{4}x+\frac{1}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{8}\right)^{2}=\frac{61}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{8}\right)^{2}}=\sqrt{\frac{61}{64}}$$

Simplify.

$$x-\frac{1}{8}=\frac{\sqrt{61}}{8} x-\frac{1}{8}=-\frac{\sqrt{61}}{8}$$

Add \frac{1}{8} to both sides of the equation.

$$x=\frac{\sqrt{61}+1}{8} x=\frac{1-\sqrt{61}}{8}$$