Solve 4(4+1)=4k^2

$$4 ( 4 + 1 ) = 4 k ^ { 2 }$$

Solve for k

$$k=\sqrt{5}\approx 2.236067977$$

$$k=-\sqrt{5}\approx -2.236067977$$

Steps by Finding Square Root

Cancel out 4 on both sides.

$$4+1=k^{2}$$

Add 4 and 1 to get 5.

$$5=k^{2}$$

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

$$k^{2}=5$$

Take the square root of both sides of the equation.

$$k=\sqrt{5} k=-\sqrt{5}$$

Steps Using the Quadratic Formula

Cancel out 4 on both sides.

$$4+1=k^{2}$$

Add 4 and 1 to get 5.

$$5=k^{2}$$

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

$$k^{2}=5$$

Subtract 5 from both sides.

$$k^{2}-5=0$$

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

$$k=\frac{0±\sqrt{0^{2}-4\left(-5\right)}}{2}$$

Square 0.

$$k=\frac{0±\sqrt{-4\left(-5\right)}}{2}$$

Multiply -4 times -5.

$$k=\frac{0±\sqrt{20}}{2}$$

Take the square root of 20.

$$k=\frac{0±2\sqrt{5}}{2}$$

Now solve the equation k=\frac{0±2\sqrt{5}}{2} when ± is plus.

$$k=\sqrt{5}$$

Now solve the equation k=\frac{0±2\sqrt{5}}{2} when ± is minus.

$$k=-\sqrt{5}$$

The equation is now solved.

$$k=\sqrt{5} k=-\sqrt{5}$$