Solve (3z-1)^2=(3z+4)(3z+5)

$$( 3 z - 1 ) ^ { 2 } = ( 3 z + 4 ) ( 3 z + 5 )$$

Solve for z

$$z=-\frac{19}{33}\approx -0.575757576$$

Steps for Solving Linear Equation

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3z-1\right)^{2}.

$$9z^{2}-6z+1=\left(3z+4\right)\left(3z+5\right)$$

Use the distributive property to multiply 3z+4 by 3z+5 and combine like terms.

$$9z^{2}-6z+1=9z^{2}+27z+20$$

Subtract 9z^{2} from both sides.

$$9z^{2}-6z+1-9z^{2}=27z+20$$

Combine 9z^{2} and -9z^{2} to get 0.

$$-6z+1=27z+20$$

Subtract 27z from both sides.

$$-6z+1-27z=20$$

Combine -6z and -27z to get -33z.

$$-33z+1=20$$

Subtract 1 from both sides.

$$-33z=20-1$$

Subtract 1 from 20 to get 19.

$$-33z=19$$

Divide both sides by -33.

$$z=\frac{19}{-33}$$

Fraction \frac{19}{-33} can be rewritten as -\frac{19}{33} by extracting the negative sign.

$$z=-\frac{19}{33}$$