Solve 2^2+3=-(sqrt{3}z-1)

$$2 ^ { 2 } + 3 = - ( \sqrt { 3 } z - 1 )$$

Solve for z

$$z=-2\sqrt{3}\approx -3.464101615$$

Steps for Solving Linear Equation

Calculate 2 to the power of 2 and get 4.

$$4+3=-\left(\sqrt{3}z-1\right)$$

Add 4 and 3 to get 7.

$$7=-\left(\sqrt{3}z-1\right)$$

To find the opposite of \sqrt{3}z-1, find the opposite of each term.

$$7=-\sqrt{3}z-\left(-1\right)$$

The opposite of -1 is 1.

$$7=-\sqrt{3}z+1$$

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

$$-\sqrt{3}z+1=7$$

Subtract 1 from both sides.

$$-\sqrt{3}z=7-1$$

Subtract 1 from 7 to get 6.

$$-\sqrt{3}z=6$$

The equation is in standard form.

$$\left(-\sqrt{3}\right)z=6$$

Divide both sides by -\sqrt{3}.

$$\frac{\left(-\sqrt{3}\right)z}{-\sqrt{3}}=\frac{6}{-\sqrt{3}}$$

Dividing by -\sqrt{3} undoes the multiplication by -\sqrt{3}.

$$z=\frac{6}{-\sqrt{3}}$$

Divide 6 by -\sqrt{3}.

$$z=-2\sqrt{3}$$